---
title: "DC Circuits: Kirchhoff’s Laws and Theorems"
book: "University Physics — Year 1"
subject: physics
language: en
chapter: 6
exercises: 12
source: https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems
---

# Chapter 6 — DC Circuits: Kirchhoff’s Laws and Theorems

Turn the key on a cold morning: the starter groans, the dashboard lights dim for a second, and the car coughs into life. The battery that delivered a steady $12.6\,\mathrm{V}$ a moment ago is down to eleven while the starter draws its hundred and fifty [amperes](#def-b1-dc-circuits-current) — and the headlights, wired in parallel, pay for it. Everything in that second is governed by two conservation laws and a handful of theorems about [linear circuits](#def-b1-dc-circuits-linear), which this chapter states, proves, and applies: [Kirchhoff’s laws](#thm-b1-dc-circuits-kirchhoff), the models of sources and [resistors](#def-b1-dc-circuits-resistor), dividers, Thévenin and Norton, and the graphical art of finding an [operating point](#met-b1-dc-circuits-loadline).

## 6.1 Current, voltage, and the quasi-steady regime

**Definition 6.1 (Electric current).**

The *electric current* through a surface (a wire’s cross-section) is the charge crossing it per unit time, counted positively in a chosen orientation:

$$
i = \frac{\dd q}{\dd t} \qquad (\text{amperes: } 1\,\mathrm{A} = 1\,\mathrm{C}/\mathrm{s}).
$$

Reversing the orientation changes the sign of $i$. In a metal the carriers are electrons, which drift *against* the conventional current direction, at a fraction of a millimeter per second.

**Definition 6.2 (Potential and voltage).**

Each point of a circuit has an *electric potential* $V$ (volts), defined up to a constant fixed by choosing a *ground* ($V = 0$). The *voltage* between $A$ and $B$ is the potential difference $u_{AB} = V_A - V_B$, drawn as an arrow from $B$ to $A$. The energy a charge $q$ loses in going from $A$ to $B$ is $q\,u_{AB}$ (this connects to the electrostatic potential of [Chapter 27](https://one-course.com/books/physics/3/en/chapter/27-electric-potential-conductors-and-capacitors#ch-b1-potential-capacitors)).

**Definition 6.3 (Quasi-steady regime (ARQS)).**

A circuit of size $\ell$ driven at frequency $f$ is in the *quasi-steady regime* when the propagation time $\ell/c$ of electrical [signals](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) along it is negligible against the period $1/f$: $\ell \ll c/f$. The current is then the same at every point of an unbranched wire at each instant, and the laws below apply at each instant to time-varying currents and [voltages](#def-b1-dc-circuits-voltage) (lowercase $i$, $u$) exactly as to steady ones ($I$, $U$).

**Example 6.4 (How quasi is quasi).**

At $50\,\mathrm{Hz}$, $c/f = 6000\,\mathrm{km}$: a house is in the [quasi-steady regime](#def-b1-dc-circuits-arqs). At $1\,\mathrm{MHz}$, $300\,\mathrm{m}$: a radio still is. At $2\,\mathrm{GHz}$, $15\,\mathrm{cm}$: a phone’s circuit board is not, and its tracks must be treated as transmission lines — a topic of the Year 2 volume.

**Theorem 6.5 (Kirchhoff’s laws).**

In the [quasi-steady regime](#def-b1-dc-circuits-arqs):

1. *[Node law](#thm-b1-dc-circuits-kirchhoff)* : at a node where several wires meet, the sum of the currents arriving equals the sum of the currents leaving, $\sum_{\text{in}} i_k = \sum_{\text{out}} i_k$ .
2. *[Loop law](#thm-b1-dc-circuits-kirchhoff)* : around any closed loop, the [voltages](#def-b1-dc-circuits-voltage) add up to zero when counted with the sign given by a chosen direction of travel: $\sum_{\text{loop}} \pm u_k = 0$ .

**Proof.** Charge is conserved and, in the [quasi-steady regime](#def-b1-dc-circuits-arqs), does not pile up at a node: what arrives per second leaves per second. The [loop law](#thm-b1-dc-circuits-kirchhoff) is the statement that the potential is a function of position: going around a loop and back to the starting point, the sum of the potential drops is $V_A - V_A = 0$. ∎

**Notation 6.6 (Conventions and power).**

For a two-terminal component (a *dipole*), the *[receiver convention](#thm-b1-dc-circuits-kirchhoff)* draws the [voltage](#def-b1-dc-circuits-voltage) arrow and the current arrow in opposite directions; the *[generator convention](#thm-b1-dc-circuits-kirchhoff)* draws them in the same direction. In the [receiver convention](#thm-b1-dc-circuits-kirchhoff), the power *received* by the dipole is

$$
p = u\,i \qquad (\text{watts}),
$$

positive for a [resistor](#def-b1-dc-circuits-resistor) that heats, negative for a battery that delivers. In the [generator convention](#thm-b1-dc-circuits-kirchhoff) $p = ui$ is the power *delivered*.

![The two sign conventions for a dipole: arrows opposed (receiver) or aligned (generator). The physics is the same; only the sign of ui is read differently.](https://one-course.com/images/onecourse/chapters/physics-3/b1-dc-circuits/fig-82a2d956fe08.svg)

*The two sign conventions for a dipole: arrows opposed (receiver) or aligned (generator). The physics is the same; only the sign of $ui$ is read differently.*

## 6.2 Dipoles and their models

**Definition 6.7 (Characteristic; resistor; Ohm’s law).**

The *characteristic* of a dipole is the curve $i(u)$ (or $u(i)$) it imposes. A *resistor* has a linear characteristic through the origin, *Ohm’s law*

$$
u = R\,i \qquad (\text{receiver convention}),
$$

$R$ being its *resistance* (ohms, $1\,\Omega = 1\,\mathrm{V}/\mathrm{A}$) and $G = 1/R$ its *conductance* (siemens). It receives $p = Ri^2 =
u^2/R \geq 0$, dissipated as heat (*Joule effect*). A wire of length $\ell$, cross-section $S$ and resistivity $\rho$ has $R = \rho\ell/S$.

**Definition 6.8 (Ideal and real sources).**

An *ideal [voltage](#def-b1-dc-circuits-voltage) source* imposes $u = E$ whatever the current; $E$ is its *electromotive force* (emf). An *ideal current source* imposes $i = I_N$ whatever the [voltage](#def-b1-dc-circuits-voltage). A *real source* (battery, generator, power supply) is modeled, in the [generator convention](#thm-b1-dc-circuits-kirchhoff), by

$$
u = E - r\,i \quad (\text{Thévenin model: ideal emf } E \text{ in series with } r)
$$

or, equivalently, by $i = I_N - u/r$ (Norton model: ideal current source $I_N = E/r$ in parallel with $r$); $r$ is the *internal resistance*. $E$ is the open-circuit [voltage](#def-b1-dc-circuits-voltage), $I_N = E/r$ the short-circuit current.

**Proposition 6.9 (Power balance of a real source).**

A real source $(E, r)$ feeding a [resistor](#def-b1-dc-circuits-resistor) $R$ delivers $i = E/(R + r)$ and the power

$$
P_R = \frac{E^2 R}{(R + r)^2},
$$

maximal and equal to $E^2/4r$ when $R = r$ (*impedance matching*), with efficiency $P_R/(Ei) = R/(R + r)$, only $50\%$ at the maximum.

**Proof.** [Loop law](#thm-b1-dc-circuits-kirchhoff): $E = ri + Ri$. $P_R = Ri^2$; differentiate $R/(R + r)^2$: $\big[(R + r)^2 - 2R(R + r)\big]/(R + r)^4 = (r - R)/(R + r)^3$, zero at $R = r$, positive before and negative after. The source supplies $Ei = (R + r)i^2$, of which $ri^2$ heats the source itself. ∎

![Left: the characteristics of a real source (u = E - ri) and of a resistive load (u = Ri) meet at the operating point. Right: the power delivered to the load peaks at R = r, where half of the source’s power is wasted in its own resistance.](https://one-course.com/images/onecourse/chapters/physics-3/b1-dc-circuits/fig-156f3e83eaa6.svg)

![Left: the characteristics of a real source (u = E - ri) and of a resistive load (u = Ri) meet at the operating point. Right: the power delivered to the load peaks at R = r, where half of the source’s power is wasted in its own resistance.](https://one-course.com/images/onecourse/chapters/physics-3/b1-dc-circuits/fig-ef081915e244.svg)

*Left: the characteristics of a real source ($u = E - ri$) and of a resistive load ($u = Ri$) meet at the [operating point](#met-b1-dc-circuits-loadline). Right: the power delivered to the load peaks at $R = r$, where half of the source’s power is wasted in its own resistance.*

**Example 6.10 (A car battery under load).**

Open circuit $12.6\,\mathrm{V}$; $11.4\,\mathrm{V}$ while the starter draws $120\,\mathrm{A}$: $r = 1.2/120 = 10\,\mathrm{m}\Omega$, $I_N = 1260\,\mathrm{A}$ (hence the welding-grade danger of a dropped spanner). Matched load $R = 10\,\mathrm{m}\Omega$ would take $E^2/4r = 4\,\mathrm{kW}$ for a few seconds — about what a starter motor needs.

**Remark 6.11 (Other dipoles).**

A *diode* conducts in one direction only: an idealized model is an open circuit for $u < U_0$ and a fixed [voltage](#def-b1-dc-circuits-voltage) $u = U_0$ (about $0.7\,\mathrm{V}$ for silicon, $2\,$ to $3\,\mathrm{V}$ for an LED) when it conducts. A filament lamp is a [resistor](#def-b1-dc-circuits-resistor) whose $R$ grows with its temperature, hence with $i$ — a curved characteristic. In steady state a capacitor is an open circuit ($i = C\,\dd u/\dd t = 0$) and an inductor a short circuit ($u = L\,\dd i/\dd t = 0$); their role in transients is [Chapter 7](https://one-course.com/books/physics/3/en/chapter/7-transient-regimes-first-and-second-order#ch-b1-transient-regimes).

## 6.3 Associations and dividers

**Proposition 6.12 (Series and parallel resistors).**

[Resistors](#def-b1-dc-circuits-resistor) in series (same current) add: $R = R_1 + R_2 + \cdots$. [Resistors](#def-b1-dc-circuits-resistor) in parallel (same [voltage](#def-b1-dc-circuits-voltage)) add their conductances: $1/R = 1/R_1 + 1/R_2 + \cdots$; for two, $R = R_1R_2/(R_1 + R_2)$, written $R_1 \parallel R_2$.

**Proof.** Series: $u = u_1 + u_2 = (R_1 + R_2)i$ ([loop law](#thm-b1-dc-circuits-kirchhoff)). Parallel: $i = i_1 +
i_2 = u/R_1 + u/R_2$ ([node law](#thm-b1-dc-circuits-kirchhoff)). ∎

**Proposition 6.13 (Voltage and current dividers).**

Two [resistors](#def-b1-dc-circuits-resistor) in series across a [voltage](#def-b1-dc-circuits-voltage) $u$ share it in proportion to their resistances:

$$
u_2 = \frac{R_2}{R_1 + R_2}\,u .
$$

Two [resistors](#def-b1-dc-circuits-resistor) in parallel fed by a current $i$ share it in inverse proportion:

$$
i_2 = \frac{R_1}{R_1 + R_2}\,i = \frac{G_2}{G_1 + G_2}\,i .
$$

**Proof.** Series: $i = u/(R_1 + R_2)$ and $u_2 = R_2 i$. Parallel: $u = (R_1
\parallel R_2)\,i$ and $i_2 = u/R_2$. ∎

![Voltage divider (left): u_2 = R_2 u/(R_1 + R_2). Current divider (right): i_2 = R_1 i/(R_1 + R_2) — the current prefers the smaller resistance.](https://one-course.com/images/onecourse/chapters/physics-3/b1-dc-circuits/fig-246b3afb158f.svg)

*[Voltage divider](#prop-b1-dc-circuits-dividers) (left): $u_2 = R_2 u/(R_1 + R_2)$. [Current divider](#prop-b1-dc-circuits-dividers) (right): $i_2 = R_1 i/(R_1 + R_2)$ — the current prefers the smaller resistance.*

**Remark 6.14 (Loading a divider).**

The divider formula holds only if nothing draws current from the midpoint. A load $R_L$ connected across $R_2$ replaces $R_2$ by $R_2
\parallel R_L$ and lowers $u_2$ — the “loading effect”, and the reason a voltmeter must have a high resistance and a potentiometer a low one compared with what it feeds ([Exercise 6.4](#exo-b1-dc-circuits-4)).

**Method 6.15 (Reducing a network).**

To find one current or [voltage](#def-b1-dc-circuits-voltage) in a network of [resistors](#def-b1-dc-circuits-resistor) and one source:

1. starting far from the source, replace series and parallel groups by their equivalents until a single [resistor](#def-b1-dc-circuits-resistor) faces the source;
2. compute the source current;
3. walk back, splitting currents with [current dividers](#prop-b1-dc-circuits-dividers) and [voltages](#def-b1-dc-circuits-voltage) with [voltage dividers](#prop-b1-dc-circuits-dividers) , down to the wanted branch.

With several sources, or for a branch whose neighbors change, use the theorems of the next section.

## 6.4 Theorems for linear circuits

**Definition 6.16 (Linear dipole, linear circuit).**

A dipole is *linear* if its characteristic is a straight line ([resistor](#def-b1-dc-circuits-resistor), ideal and real sources); a circuit made only of linear dipoles is a *linear circuit*. Its unknowns obey a system of linear equations ([Kirchhoff’s laws](#thm-b1-dc-circuits-kirchhoff) plus the characteristics), whose solution is unique.

**Theorem 6.17 (Thévenin and Norton).**

Seen from two of its terminals $A$ and $B$, any [linear circuit](#def-b1-dc-circuits-linear) is equivalent to a real source: an emf $E_{\mathrm{Th}}$ in series with a resistance $R_{\mathrm{Th}}$ (Thévenin), or a [current source](#def-b1-dc-circuits-sources) $I_N =
E_{\mathrm{Th}}/R_{\mathrm{Th}}$ in parallel with $R_{\mathrm{Th}}$ (Norton), where

- $E_{\mathrm{Th}}$ is the open-circuit [voltage](#def-b1-dc-circuits-voltage) $u_{AB}$ ;
- $I_N$ is the short-circuit current from $A$ to $B$ ;
- $R_{\mathrm{Th}}$ is the resistance seen between $A$ and $B$ when every independent source is switched off (ideal [voltage sources](#def-b1-dc-circuits-sources) replaced by wires, ideal [current sources](#def-b1-dc-circuits-sources) by open circuits).

**Partial proof.** By linearity, the relation between the [voltage](#def-b1-dc-circuits-voltage) $u_{AB}$ and the current $i$ drawn at the terminals is affine: $u_{AB} = a - b\,i$. Setting $i = 0$ identifies $a = E_{\mathrm{Th}}$; setting $u_{AB} = 0$ gives $b = E_{\mathrm{Th}}/I_N$. That $b$ equals the resistance seen with sources killed follows from superposition (below): the part of $u_{AB}$ due to the external current alone is what a passive network of resistance $R_{\mathrm{Th}}$ would produce. That every linear system has an affine input–output relation is the algebra of linear equations. ∎

**Theorem 6.18 (Superposition).**

In a [linear circuit](#def-b1-dc-circuits-linear) with several independent sources, any current or [voltage](#def-b1-dc-circuits-voltage) is the sum of the values it takes when each source acts alone, the others being switched off.

**Proof.** The equations are linear in the unknowns, with the sources as the right-hand side; the solution is a linear function of that right-hand side, hence the sum of the solutions for each source separately. ∎

**Proposition 6.19 (Millman’s theorem).**

A node $N$ connected to points of potentials $V_1, \dots, V_n$ through resistances $R_1, \dots, R_n$ (and to nothing else) has the potential

$$
V_N = \frac{\sum_k V_k/R_k}{\sum_k 1/R_k} = \frac{\sum_k G_k V_k}{\sum_k G_k},
$$

the conductance-weighted mean of its neighbors’ potentials.

**Proof.** [Node law](#thm-b1-dc-circuits-kirchhoff) at $N$: $\sum_k (V_k - V_N)/R_k = 0$; solve for $V_N$. ∎

![Two sources feeding a common load R_3. Seen from R_3’s terminals, the rest of the circuit is a Thévenin source (E_ Th, R_ Th); Millman’s theorem gives the potential of node N in one line.](https://one-course.com/images/onecourse/chapters/physics-3/b1-dc-circuits/fig-30434c8dc9c6.svg)

*Two sources feeding a common load $R_3$. Seen from $R_3$’s terminals, the rest of the circuit is a Thévenin source ($E_{\mathrm{Th}}$, $R_{\mathrm{Th}}$); Millman’s theorem gives the potential of node $N$ in one line.*

**Example 6.20 (Three ways to the same answer).**

In the figure, $E_1 = 10\,\mathrm{V}$, $R_1 = 1.0\,\mathrm{k}\Omega$, $E_2 =
5.0\,\mathrm{V}$, $R_2 = 2.0\,\mathrm{k}\Omega$, $R_3 = 2.0\,\mathrm{k}\Omega$. *Millman*: $V_N = (10/1 + 5/2 + 0/2)/(1 + 0.5 + 0.5) = 12.5/2 =
6.25\,\mathrm{V}$. *Superposition*: $E_1$ alone sees $R_2 \parallel
R_3 = 1\,\mathrm{k}\Omega$, so $u = 10 \times 1/2 = 5\,\mathrm{V}$; $E_2$ alone sees $R_1 \parallel R_3 = 0.667\,\mathrm{k}\Omega$, $u = 5 \times 0.667/2.667
= 1.25\,\mathrm{V}$; total $6.25\,\mathrm{V}$. *Thévenin* seen from $R_3$: $E_{\mathrm{Th}}$ is the open-circuit [voltage](#def-b1-dc-circuits-voltage) of the divider $(E_1, R_1, R_2, E_2)$: $E_2 + (E_1 - E_2)R_2/(R_1 + R_2) = 5 + 5 \times
2/3 = 8.33\,\mathrm{V}$; $R_{\mathrm{Th}} = R_1 \parallel R_2 =
0.667\,\mathrm{k}\Omega$; then $u = 8.33 \times 2/2.667 = 6.25\,\mathrm{V}$.

## 6.5 Operating point; the Wheatstone bridge

**Method 6.21 (Graphical operating point).**

A non-linear dipole (characteristic $i = g(u)$) fed by a Thévenin source $(E, R)$ settles where both relations hold: draw the dipole’s characteristic and the source’s *[load line](#met-b1-dc-circuits-loadline)* $u = E - Ri$ on the same axes; their intersection is the *[operating point](#met-b1-dc-circuits-loadline)*. Designing the series [resistor](#def-b1-dc-circuits-resistor) of an LED, a Zener regulator or a transistor’s bias is exactly this construction.

![An LED (threshold near 2\, V) fed by E = 5\, V through R = 200\,: the load line u = E - Ri crosses the diode’s characteristic at 2.0\, V, 15\, mA. Raising R tilts the line down and dims the LED.](https://one-course.com/images/onecourse/chapters/physics-3/b1-dc-circuits/fig-cc3d03136dcd.svg)

*An LED (threshold near $2\,\mathrm{V}$) fed by $E = 5\,\mathrm{V}$ through $R = 200\,\Omega$: the [load line](#met-b1-dc-circuits-loadline) $u = E - Ri$ crosses the diode’s characteristic at $2.0\,\mathrm{V}$, $15\,\mathrm{mA}$. Raising $R$ tilts the line down and dims the LED.*

**Proposition 6.22 (Wheatstone bridge).**

Four [resistors](#def-b1-dc-circuits-resistor) $R_1, R_2$ (one branch) and $R_3, R_4$ (the other) across a source $E$; the bridge output is the [voltage](#def-b1-dc-circuits-voltage) between the two midpoints:

$$
u = E\left(\frac{R_2}{R_1 + R_2} - \frac{R_4}{R_3 + R_4}\right),
$$

zero (the bridge is *balanced*) iff $R_1R_4 = R_2R_3$. For a balanced bridge with $R_4$ perturbed to $R_4(1 + \epsilon)$, $\epsilon
\ll 1$, and $R_3 = R_4$: $u \approx -E\epsilon/4$.

**Proof.** Two [voltage dividers](#prop-b1-dc-circuits-dividers); subtract. Balance: $R_2(R_3 + R_4) = R_4(R_1 +
R_2)$, i.e. $R_2R_3 = R_1R_4$. With $R_3 = R_4 = R$ and $R_1 = R_2$: $u = E[\tfrac12 - (1 + \epsilon)/(2 + \epsilon)] = E[(2 + \epsilon - 2 -
2\epsilon)/(2(2 + \epsilon))] \approx -E\epsilon/4$. ∎

![The Wheatstone bridge: two dividers side by side. Balanced when R_1R_4 = R_2R_3, it converts a tiny change of one resistance (strain gauge, thermistor) into a voltage around zero, which is far easier to amplify than a small change of a large voltage.](https://one-course.com/images/onecourse/chapters/physics-3/b1-dc-circuits/fig-bef8e1df3350.svg)

*The [Wheatstone bridge](#prop-b1-dc-circuits-wheatstone): two dividers side by side. Balanced when $R_1R_4 = R_2R_3$, it converts a tiny change of one resistance (strain gauge, thermistor) into a [voltage](#def-b1-dc-circuits-voltage) around zero, which is far easier to amplify than a small change of a large [voltage](#def-b1-dc-circuits-voltage).*

## 6.6 Exercises

**Exercise 6.1 ★.**

A phone charges at $1.5\,\mathrm{A}$ for one hour: what charge flows, how many electrons? Estimate whether the [quasi-steady regime](#def-b1-dc-circuits-arqs) holds for the house wiring at $50\,\mathrm{Hz}$, for a $10\,\mathrm{cm}$ circuit at $1\,\mathrm{MHz}$, and for the same circuit at $1\,\mathrm{GHz}$.

**Solution of Exercise 6.1.**

$q = 1.5 \times 3600 = 5400\,\mathrm{C}$, $N = 5400/1.6\times10^{-19} =
3.4 \times 10^{22}$ electrons. $c/f$: $6000\,\mathrm{km}$ at $50\,\mathrm{Hz}$ (house: yes); $300\,\mathrm{m}$ at $1\,\mathrm{MHz}$ ($10\,\mathrm{cm}$ circuit: yes); $30\,\mathrm{cm}$ at $1\,\mathrm{GHz}$, comparable to $10\,\mathrm{cm}$: no.

**Exercise 6.2 ★.**

A $2000\,\mathrm{W}$ heater on $230\,\mathrm{V}$: current, resistance, energy used in $3.0\,\mathrm{h}$ (in kWh and in joules). Its $5.0\,\mathrm{m}$ copper cord has a cross-section of $1.5\,\mathrm{mm}^{2}$ ($\rho = 1.7 \times 10^{-8}\,\Omega\,\mathrm{m}$): power lost in the cord.

**Solution of Exercise 6.2.**

$I = 2000/230 = 8.7\,\mathrm{A}$; $R = U^2/P = 26.5\,\Omega$; $E =
6.0\,\mathrm{kWh} = 2.2 \times 10^{7}\,\mathrm{J}$. Cord: one conductor $R = 1.7\times10^{-8}
\times 5.0/1.5\times10^{-6} = 57\,\mathrm{m}\Omega$, two conductors $0.11\,\Omega$: $P = 0.11 \times 8.7^2 = 8.6\,\mathrm{W}$.

**Exercise 6.3 ★.**

With $10\,\Omega$, $20\,\Omega$ and $30\,\Omega$: compute the resistance of the three in series, in parallel, and of $(10 \parallel 20) + 30$.

**Solution of Exercise 6.3.**

Series $60\,\Omega$; parallel $1/(0.1 + 0.05 + 0.033) = 5.5\,\Omega$; $10 \parallel 20 = 6.7\,\Omega$, plus $30$: $36.7\,\Omega$.

**Exercise 6.4 ★.**

A $12\,\mathrm{V}$ source feeds $4.7\,\mathrm{k}\Omega$ and $2.2\,\mathrm{k}\Omega$ in series. Compute the [voltage](#def-b1-dc-circuits-voltage) across the $2.2\,\mathrm{k}\Omega$. A device of resistance $2.2\,\mathrm{k}\Omega$ is then connected across it: new [voltage](#def-b1-dc-circuits-voltage)? Conclude on the loading effect.

**Solution of Exercise 6.4.**

$u_2 = 12 \times 2.2/6.9 = 3.83\,\mathrm{V}$. Loaded: $2.2 \parallel 2.2 =
1.1\,\mathrm{k}\Omega$, $u_2 = 12 \times 1.1/5.8 = 2.28\,\mathrm{V}$: the load pulls the divider down by $40\%$ — a divider only “divides” if it feeds something of much higher resistance.

**Exercise 6.5 ★★.**

A battery reads $12.6\,\mathrm{V}$ open and $11.4\,\mathrm{V}$ while delivering $120\,\mathrm{A}$. Find its [internal resistance](#def-b1-dc-circuits-sources) and short-circuit current, the power delivered to the starter and the power lost inside, and the efficiency.

**Solution of Exercise 6.5.**

$r = (12.6 - 11.4)/120 = 10\,\mathrm{m}\Omega$; $I_N = 12.6/0.010 =
1260\,\mathrm{A}$. To the starter $11.4 \times 120 = 1.37\,\mathrm{kW}$; lost $0.010 \times 120^2 = 144\,\mathrm{W}$; efficiency $1368/1512 = 90\%$.

**Exercise 6.6 ★★.**

Find the Thévenin equivalent, seen from the $2.2\,\mathrm{k}\Omega$ [resistor](#def-b1-dc-circuits-resistor)’s terminals, of the divider of [Exercise 6.4](#exo-b1-dc-circuits-4) ($E = 12\,\mathrm{V}$, $R_1 = 4.7\,\mathrm{k}\Omega$, $R_2 = 2.2\,\mathrm{k}\Omega$). Use it to recover the loaded [voltage](#def-b1-dc-circuits-voltage) of that exercise.

**Solution of Exercise 6.6.**

$E_{\mathrm{Th}} = 3.83\,\mathrm{V}$ (open-circuit divider); $R_{\mathrm{Th}}
= 4.7 \parallel 2.2 = 1.50\,\mathrm{k}\Omega$ (source shorted). Loaded: $u = 3.83 \times 2.2/(1.50 + 2.2) = 2.28\,\mathrm{V}$.

**Exercise 6.7 ★★.**

Two batteries, $E_1 = 10\,\mathrm{V}$ ($r_1 = 1.0\,\Omega$) and $E_2 = 5.0\,\mathrm{V}$ ($r_2 = 2.0\,\Omega$), are connected in parallel (same polarity) across a $2.0\,\Omega$ load. Find the load [voltage](#def-b1-dc-circuits-voltage) by superposition, then by Millman, and the current in each battery. Is the weaker battery charging or discharging?

**Solution of Exercise 6.7.**

Superposition: $E_1$ alone sees $r_2 \parallel R = 1.0\,\Omega$, $u = 10 \times 1/(1 + 1) = 5.0\,\mathrm{V}$; $E_2$ alone sees $r_1 \parallel R
= 0.667\,\Omega$, $u = 5 \times 0.667/2.667 = 1.25\,\mathrm{V}$; total $6.25\,\mathrm{V}$. Millman: $(10/1 + 5/2 + 0/2)/(1 + 0.5 + 0.5) = 6.25\,\mathrm{V}$. Battery 1: $(10 - 6.25)/1 = 3.75\,\mathrm{A}$ out; battery 2: $(5 - 6.25)/2
= -0.63\,\mathrm{A}$, i.e. $0.63\,\mathrm{A}$ *into* it — it is being charged by the stronger one; load $6.25/2 = 3.1\,\mathrm{A}$.

**Exercise 6.8 ★★.**

A source $E = 9.0\,\mathrm{V}$, $r = 3.0\,\Omega$. Compute the power delivered to loads of $1.0\,\Omega$, $3.0\,\Omega$, $12\,\Omega$, and the efficiency in each case. Which load would an engineer choose for a battery-powered device, and why not the matched one?

**Solution of Exercise 6.8.**

$i = 9/(R + 3)$, $P = Ri^2$, $\eta = R/(R + 3)$: $1\,\Omega$: $5.1\,\mathrm{W}$, $25\%$; $3\,\Omega$: $6.75\,\mathrm{W}$, $50\%$; $12\,\Omega$: $4.3\,\mathrm{W}$, $80\%$. A battery device wants efficiency (battery life), hence $R \gg r$; the matched load wastes half the energy as heat in the battery.

**Exercise 6.9 ★★.**

An LED conducts at $2.0\,\mathrm{V}$ and must carry $15\,\mathrm{mA}$ from a $5.0\,\mathrm{V}$ supply. Choose the series [resistor](#def-b1-dc-circuits-resistor), the powers in the [resistor](#def-b1-dc-circuits-resistor) and in the LED, and sketch the [load line](#met-b1-dc-circuits-loadline) construction. What happens if the supply is $3.3\,\mathrm{V}$?

**Solution of Exercise 6.9.**

$R = (5.0 - 2.0)/0.015 = 200\,\Omega$; $P_R = 3.0 \times 0.015 =
45\,\mathrm{mW}$, $P_{\mathrm{LED}} = 2.0 \times 0.015 = 30\,\mathrm{mW}$. [Load line](#met-b1-dc-circuits-loadline) from $(0, 25\,\mathrm{mA})$ to $(5\,\mathrm{V}, 0)$ meets the vertical characteristic at $2.0\,\mathrm{V}$. At $3.3\,\mathrm{V}$ the same [resistor](#def-b1-dc-circuits-resistor) gives only $6.5\,\mathrm{mA}$; for $15\,\mathrm{mA}$ one needs $R = 1.3/0.015 =
87\,\Omega$.

**Exercise 6.10 ★★★.**

A strain gauge $R_4 = R(1 + \epsilon)$, $R = 1.0\,\mathrm{k}\Omega$, sits in a [Wheatstone bridge](#prop-b1-dc-circuits-wheatstone) with three fixed $1.0\,\mathrm{k}\Omega$ [resistors](#def-b1-dc-circuits-resistor) across $5.0\,\mathrm{V}$. Derive the output [voltage](#def-b1-dc-circuits-voltage) to first order in $\epsilon$ and compute it for $\epsilon = 1.0 \times 10^{-3}$. Why is the bridge preferable to measuring the gauge’s [voltage](#def-b1-dc-circuits-voltage) in a simple divider?

**Solution of Exercise 6.10.**

$u = E\left[\dfrac12 - \dfrac{1 + \epsilon}{2 + \epsilon}\right] =
-E\dfrac{\epsilon}{2(2 + \epsilon)} \approx -\dfrac{E\epsilon}{4} =
-1.25\,\mathrm{mV}$. In a simple divider the same change sits on top of a $2.5\,\mathrm{V}$ offset ($2.501\,25\,\mathrm{V}$); the bridge delivers the $1.25\,\mathrm{mV}$ around zero, which can be amplified a thousandfold without saturating anything.

**Exercise 6.11 ★★★.**

Twelve identical [resistors](#def-b1-dc-circuits-resistor) $R$ form the edges of a cube. Using symmetry (which nodes are at the same potential when a current $I$ enters one corner and leaves the opposite one), show that the resistance between opposite corners is $5R/6$.

**Solution of Exercise 6.11.**

By symmetry the three neighbors of the entry corner share one potential, the three neighbors of the exit corner another. The current $I$ splits into $I/3$ in the three entry edges, each then into two $I/6$ along the six middle edges, which recombine into $I/3$ in the three exit edges. [Voltage](#def-b1-dc-circuits-voltage): $RI/3 + RI/6 + RI/3 = 5RI/6$, so $R_{\mathrm{eq}} = 5R/6$.

**Exercise 6.12 ★★★.**

A [current source](#def-b1-dc-circuits-sources) $I_N = 2.0\,\mathrm{A}$ in parallel with $4.0\,\Omega$ is connected, through a $2.0\,\Omega$ [resistor](#def-b1-dc-circuits-resistor), to a $6.0\,\mathrm{V}$ battery (negligible resistance) whose emf opposes it. Convert the Norton source to Thévenin and find the current in the $2.0\,\Omega$ [resistor](#def-b1-dc-circuits-resistor) and the power exchanged by each source.

**Solution of Exercise 6.12.**

Thévenin: $E = I_N r = 8.0\,\mathrm{V}$ with $4.0\,\Omega$ in series. Loop: $8.0 - 6.0 = (4.0 + 2.0)i$, $i = 0.33\,\mathrm{A}$ toward the battery. Battery receives $6.0 \times 0.33 = 2.0\,\mathrm{W}$ (charging); the $2\,\Omega$ [resistor](#def-b1-dc-circuits-resistor) dissipates $0.22\,\mathrm{W}$; the terminal [voltage](#def-b1-dc-circuits-voltage) of the Norton pair is $8.0 - 4.0 \times 0.33 = 6.67\,\mathrm{V}$, so its $4\,\Omega$ carries $1.67\,\mathrm{A}$ and dissipates $11.1\,\mathrm{W}$, and the [current source](#def-b1-dc-circuits-sources) delivers $2.0 \times 6.67 = 13.3\,\mathrm{W}$ — which adds up.

![Under the bonnet: the 12\, V battery, its thick cables and the alternator — the direct-current network of the weekend problem, with hundreds of amperes at start-up.](https://one-course.com/images/onecourse/chapters/physics-3/b1-dc-circuits/img-bd8366c9fe62.jpg)

*Under the bonnet: the $12\,\mathrm{V}$ battery, its thick cables and the alternator — the direct-current network of the weekend problem, with hundreds of [amperes](#def-b1-dc-circuits-current) at start-up.*

## 6.7 Problem: A car’s electrical system

**Problem 6.1.**

Weekend problem — one battery, a starter, two headlights, an alternator and a few meters of copper: who gets the current, who pays, and why the lights dim when the engine cranks

The battery is a real source, emf $E = 12.6\,\mathrm{V}$, [internal resistance](#def-b1-dc-circuits-sources) $r = 10\,\mathrm{m}\Omega$. Copper: $\rho = 1.7 \times 10^{-8}\,\Omega\,\mathrm{m}$.

**Part I — The battery alone.**

1. Draw the [Thévenin model](#def-b1-dc-circuits-sources) and write $u(i)$ in the [generator convention](#thm-b1-dc-circuits-kirchhoff) .
2. Give the open-circuit [voltage](#def-b1-dc-circuits-voltage) and the short-circuit current. Why must a dropped wrench never bridge the terminals?
3. Write the [Norton model](#def-b1-dc-circuits-sources) of the same battery.
4. The battery feeds a [resistor](#def-b1-dc-circuits-resistor) $R$ : express the current, the terminal [voltage](#def-b1-dc-circuits-voltage) , the power $P_R$ delivered and the power lost in $r$ .
5. Show that $P_R$ is maximal for $R = r$ and compute that maximum. What is the efficiency then?
6. A technician measures $u = 12.48\,\mathrm{V}$ at $12\,\mathrm{A}$ and $u = 11.40\,\mathrm{V}$ at $120\,\mathrm{A}$ . Check that both points lie on the model and explain how a series of such points gives $E$ and $r$ by a straight-line fit.
7. The battery stores $60\,\mathrm{A}\,\mathrm{h}$ . How long could it run the two $55\,\mathrm{W}$ headlights alone (take $12\,\mathrm{V}$ )? How much energy is that in joules?

**Part II — Cranking with the lights on.** The starter is a [resistor](#def-b1-dc-circuits-resistor) $R_s = 80\,\mathrm{m}\Omega$; each headlight is rated $55\,\mathrm{W}$ at $12.0\,\mathrm{V}$ and is treated as a [resistor](#def-b1-dc-circuits-resistor).

8. Compute the resistance of one headlight and of the two in parallel.
9. Starter alone: current, terminal [voltage](#def-b1-dc-circuits-voltage) , power in the starter, power lost in the battery.
10. Starter and both headlights together: equivalent load resistance, total current, terminal [voltage](#def-b1-dc-circuits-voltage) .
11. Use the [current divider](#prop-b1-dc-circuits-dividers) to find the starter current and the headlight current.
12. Compute the power each headlight now receives and compare with its rating: by what fraction do the lights dim?
13. Explain in two sentences, using the [loop law](#thm-b1-dc-circuits-kirchhoff) , why the lights dim precisely because the starter draws a large current.
14. Would thicker battery cables help the lights? Justify qualitatively now (Part IV quantifies).

**Part III — Engine running: the alternator.** The alternator is a second real source, $E_a = 14.2\,\mathrm{V}$, $r_a = 50\,\mathrm{m}\Omega$, in parallel with the battery; the starter is off and both headlights are on.

15. Draw the circuit: two real sources and the headlight load in parallel between the same two nodes.
16. Apply Millman’s theorem to find the common terminal [voltage](#def-b1-dc-circuits-voltage) .
17. Deduce the alternator current, the battery current (sign!) and the headlight current; check the [node law](#thm-b1-dc-circuits-kirchhoff) .
18. Is the battery charging or discharging? What power does it receive, and what power does the alternator supply?
19. Recover the terminal [voltage](#def-b1-dc-circuits-voltage) by superposition (each source alone, the other replaced by its [internal resistance](#def-b1-dc-circuits-sources) ) and check.

**Part IV — Sensors and wires.**

20. The coolant sensor is a thermistor whose resistance falls from $2.0\,\mathrm{k}\Omega$ at $20{}^{\circ}\mathrm{C}$ to $300\,\Omega$ at $90{}^{\circ}\mathrm{C}$ ; it is the lower [resistor](#def-b1-dc-circuits-resistor) of a divider fed by $5.0\,\mathrm{V}$ , the upper [resistor](#def-b1-dc-circuits-resistor) being $R_0$ . Express the output [voltage](#def-b1-dc-circuits-voltage) , and compute it at both temperatures for $R_0 = 775\,\Omega$ .
21. Show that the output swing between the two temperatures is largest when $R_0$ is the geometric mean of the two sensor values, and check the value above.
22. The same sensor is put in a [Wheatstone bridge](#prop-b1-dc-circuits-wheatstone) with three $775\,\Omega$ [resistors](#def-b1-dc-circuits-resistor) : write the bridge output and say at what temperature it is balanced.
23. The starter cable is $1.5\,\mathrm{m}$ long each way. Compute the resistance of one conductor of cross-section $16\,\mathrm{mm}^{2}$ , the total [voltage](#def-b1-dc-circuits-voltage) drop at $140\,\mathrm{A}$ , and the power dissipated in the cables.
24. Same with $4\,\mathrm{mm}^{2}$ cable: why is starter cable so thick?
25. Summarize: give the terminal [voltage](#def-b1-dc-circuits-voltage) during cranking, the fraction of power reaching the starter, and the one law that explains the dimming of the lights.

**Solution of Problem 6.1.**

**1.** Ideal emf $E$ in series with $r$: $u = E - ri$.

**2.** $u(0) = 12.6\,\mathrm{V}$; $I_N = E/r = 1260\,\mathrm{A}$. A wrench across the terminals would carry it: $16\,\mathrm{kW}$ in a few grams of steel — it melts and sprays.

**3.** [Current source](#def-b1-dc-circuits-sources) $I_N = 1260\,\mathrm{A}$ in parallel with $10\,\mathrm{m}\Omega$.

**4.** $i = E/(R + r)$; $u = ER/(R + r)$; $P_R = E^2R/(R + r)^2$; $P_r = E^2r/(R + r)^2$.

**5.** $\dd P_R/\dd R \propto (r - R)$: maximum at $R = r$, $P_{\max} = E^2/4r = 12.6^2/0.040 = 4.0\,\mathrm{kW}$; efficiency $R/(R + r) = 50\%$.

**6.** $12.6 - 0.010 \times 12 = 12.48\,\mathrm{V}$; $12.6 - 0.010 \times
120 = 11.40\,\mathrm{V}$. Plotting $u$ against $i$ gives a straight line of intercept $E$ and slope $-r$ ([least squares](https://one-course.com/books/physics/3/en/chapter/1-units-dimensions-and-measurement#prop-b1-units-dimensions-regression), [Chapter 1](https://one-course.com/books/physics/3/en/chapter/1-units-dimensions-and-measurement#ch-b1-units-dimensions)).

**7.** Two lamps: $110\,\mathrm{W}$ at $12\,\mathrm{V}$, $9.2\,\mathrm{A}$; $60/9.2 = 6.5\,\mathrm{h}$. Energy $60 \times 3600 \times 12 = 2.6\,\mathrm{MJ}$.

**8.** $R_h = 12^2/55 = 2.62\,\Omega$; in parallel $1.31\,\Omega$.

**9.** $i = 12.6/0.090 = 140\,\mathrm{A}$; $u = 12.6 - 1.4 = 11.2\,\mathrm{V}$; $P_s = 0.080 \times 140^2 = 1.57\,\mathrm{kW}$; lost $0.010 \times 140^2 =
196\,\mathrm{W}$.

**10.** $R = 0.080 \parallel 1.31 = 75.4\,\mathrm{m}\Omega$; $i = 12.6/0.0854
= 148\,\mathrm{A}$; $u = 12.6 - 1.48 = 11.1\,\mathrm{V}$.

**11.** $i_s = i \times 1.31/(1.31 + 0.080) = 139\,\mathrm{A}$; $i_h =
8.5\,\mathrm{A}$ for the two lamps.

**12.** Each lamp: $u^2/R_h = 11.1^2/2.62 = 47\,\mathrm{W}$ against $55\,\mathrm{W}$: down by $14\%$.

**13.** [Loop law](#thm-b1-dc-circuits-kirchhoff): the terminal [voltage](#def-b1-dc-circuits-voltage) is $u = E - ri$; the starter’s $140\,\mathrm{A}$ produces a $1.4\,\mathrm{V}$ drop inside the battery, and the lamps, in parallel on the same terminals, receive that reduced [voltage](#def-b1-dc-circuits-voltage).

**14.** Only partly: thicker cables cut the drop in the cables, not in $r$; lamps fed from the battery terminals still see $E - ri$.

**15.** Three branches between the same two nodes: $(E, r)$, $(E_a, r_a)$ and $R_h = 1.31\,\Omega$.

**16.** $V = \dfrac{12.6/0.010 + 14.2/0.050 + 0/1.31}{100 + 20 + 0.763}
= \dfrac{1544}{120.8} = 12.79\,\mathrm{V}$.

**17.** Alternator $(14.2 - 12.79)/0.050 = 28.3\,\mathrm{A}$; battery $(12.6 - 12.79)/0.010 = -18.6\,\mathrm{A}$ (flowing *into* it); lamps $12.79/1.31 = 9.8\,\mathrm{A}$; $28.3 = 18.6 + 9.8$.

**18.** Charging. It receives $12.79 \times 18.6 = 238\,\mathrm{W}$ ($234\,\mathrm{W}$ stored, $3.5\,\mathrm{W}$ heat); the alternator supplies $12.79 \times 28.3 = 362\,\mathrm{W}$ at its terminals.

**19.** Battery alone (alternator $\to$ $50\,\mathrm{m}\Omega$): load $0.050 \parallel 1.31 = 48.2\,\mathrm{m}\Omega$, $u_1 = 12.6 \times 48.2/58.2 =
10.44\,\mathrm{V}$. Alternator alone (battery $\to$ $10\,\mathrm{m}\Omega$): load $0.010 \parallel 1.31 = 9.92\,\mathrm{m}\Omega$, $u_2 = 14.2 \times 9.92/59.9 =
2.35\,\mathrm{V}$. Sum $12.79\,\mathrm{V}$.

**20.** $u = 5.0\,R_s/(R_0 + R_s)$: $3.60\,\mathrm{V}$ at $20{}^{\circ}\mathrm{C}$, $1.40\,\mathrm{V}$ at $90{}^{\circ}\mathrm{C}$.

**21.** $\Delta u = 5[R_a/(R_0 + R_a) - R_b/(R_0 + R_b)]$; $\dd\Delta u/\dd R_0 = 0 \Leftrightarrow R_a/(R_0 + R_a)^2 = R_b/(R_0 +
R_b)^2 \Leftrightarrow R_0^2 = R_aR_b$: $R_0 = \sqrt{2000 \times 300} =
775\,\Omega$.

**22.** $u = 5[\tfrac12 - R_s/(775 + R_s)]$, zero when $R_s =
775\,\Omega$, i.e. at the mid-range temperature where the thermistor reaches that value.

**23.** $R = 1.7\times10^{-8} \times 1.5/16\times10^{-6} =
1.6\,\mathrm{m}\Omega$ per conductor, $3.2\,\mathrm{m}\Omega$ total; drop $0.0032
\times 140 = 0.45\,\mathrm{V}$; power $63\,\mathrm{W}$.

**24.** $4\,\mathrm{mm}^{2}$: $6.4\,\mathrm{m}\Omega$ each, $12.8\,\mathrm{m}\Omega$ total: drop $1.8\,\mathrm{V}$, $250\,\mathrm{W}$ wasted — the starter would lose a sixth of its [voltage](#def-b1-dc-circuits-voltage). Thick cable keeps $R_{\text{cable}} \ll R_s$.

**25.** About $11.1\,\mathrm{V}$ at the terminals during cranking; the starter receives $1568/(1568 + 196) \approx 89\%$ of the battery’s power; the dimming is the [loop law](#thm-b1-dc-circuits-kirchhoff), $u = E - ri$.