Batteries; Kirchhoff’s rules


My notes of the excellent lectures 10 and 12 by “Walter Lewin. 8.02 Electricity and Magnetism. Spring 2002. The material is neither affiliated with nor endorsed by MIT, License: Creative Commons BY-NC-SA.”


Power supplies maintain a constant potential difference.

In the power supply the electric field opposes the current. Some kind of pump mechanism must force the current to go inside the supply against the electric field.

In the case of common batteries, it is chemical energy that provides the energy.

Zinc-Copper battery

A zinc and copper plate in a solution of H2SO4

You will get a potential difference between the two plates. To understand that takes quantum mechanics. Instead we will try to do it on a napkin.

The solution is not the key, because if you take two different conductors and touch them together, there will also be a potential difference.

The ions can flow freely from one side to the other. We connect the plates with a resistor, so a current will flow. The electric field must be from plus to minus, so inside the battery it goes against the current.

On the right copper plate, you have SO4 and CU++ ions. On the left zinc plate, you have Zn++SO4 ions

  1. On the right, as current start to run, the current carrier SO4, go from the right to the left. Doing so, they travel through an electric field that opposes their motion. Doing so, they engage in a chemical reaction that yields more energy than it costs to go against the electric field.
  2. With the SO4 current is flowing from the right to the left, there are fewer SO4 ions at the right. With the liquid remaining neutral, the Cu++ must disappear, and precipitates onto the copper plate.
  3. On the left side, you get an increase of SO4. For the liquid to remain neutral, you must also get an increase of Zn++. Some of the zinc will dissolve, to increase the concentration of the zinc ions.

So, the charge carriers in this battery, the SO4 ions, travel through an electric field that opposes their motion. This happens at the expense of chemical energy.

When the copper solution becomes very dilute, as all the copper has been plated onto the copper plate, and the left side becomes concentrated Zn++, then the battery stops.

Now you can force a current to run in the opposite direction using another external power supply, and the chemical reactions will reverse. Copper will go back in the solution, and the zinc will be precipitated onto the zinc plate. Then you can run the battery again.

A car battery is this kind of battery, except that you have lead and lead oxide instead of zinc and copper, but you also have sulfuric acid. Another well know battery is nickel-cadmium.


We will be using the following symbol for batteries

Instead of \(\Delta V\), we use the European symbol
for voltage difference: \(U\).

Internal resistance

In reality, batteries are not ideal, and have an internal resistance \(r_i\) as shown below

If the current starts to flow, it goes through \(r_i\) and \(R\). Unloaded \(R=\infty\), the external battery voltage is the same as the electromotive force (EMF), \(\mathcal{E}\) $$ V_b = \mathcal{E} $$

When you connect resistor \(R\), a current \(I\) will and the external battery voltage follows from Ohm’s law $$ \begin{align*} \mathcal{E} &= I\,(r_i + R) \\ \Rightarrow I &= \frac{\mathcal{E}}{r_i + R} & (V_b = I\,R) \\ \Rightarrow V_b &= \mathcal{E}\frac{R}{r_i + R} \end{align*} $$

In series

You can put batteries in series, to get a higher potential difference.

That brings the potential difference of the open circuit to \(2\,\mathcal{E}\).


If a charge \(dq\) moves from point A to point B,

With \(V_A \gt V_B\), the electric field is from \(A\) to \(B\).

I move charge from \(A\) to \(B\), then the electric field is doing work \(dW\) $$ dW = dq (V_A – V_B) \nonumber $$

The work is positive, if the charge is positive, or negative if the charge is negative. Take the derivative in time on both sides $$ \frac{dW}{dt} = \frac{dq}{dt} (V_A – V_B) \nonumber $$

On the left side, we now have work per unit time. That is power in Joules/s. On the right side, \(\frac{dq}{dt}\) is current in Coulombs/s. The potential difference we simply call \(V\) $$ \shaded{ P = I\,V } \quad \left[\rm W =\frac{\rm J}{\rm s} \right] \tag{Power} \label{eq:power} $$

If you can use Ohm’s law (\(V=I\,R\)), you can write this as $$ \shaded{ P = I^2\,R = \frac{V^2}{R} } \quad \left[\rm W =\frac{\rm J}{\rm s} \right] \tag{Power} \label{eq:powerohmslaw} $$

Instead of writing \({\rm J}/{\rm s}\), we write \(\rm W\) for Watt, named after the 18th-century Scottish inventor James Watt.



Suppose we have resistance \(R=100\,\Omega\) and we run a current \(I=1\,\rm A\) through it,

Then the power dissipated in this resistor, provided by the battery $$ P = I^2\, R = 1^2(100) = 1000\,\rm W \nonumber $$

The energy is dissipated as heat. If it gets hot enough, the resistor might produce light. E.g. the filament in a tungsten light bulb heats up to \(\approx 2500\,^{\circ}\rm C\) and emits light. But the amount of light is produces is no more that 20% of the power it dissipates.


If you take a 9 volt battery with an internal resistance of about 2 ohms, and short it

The power dissipated inside the battery is $$ P_{max} = \frac{V^2}{r_i} = \frac{9^2}{2} = 40\,\rm W \nonumber $$


If you take a 12 volt car battery with an internal resistance of about 0.02 ohms, and short it, the power dissipated inside the battery is $$ P_{max} = \frac{V^2}{r_i} = \frac{12^2}{0.02} = 7000\,\rm {W} \nonumber $$

The sulfuric acid will boil, and the case may melt. This is very dangerous.


The electric utility company charges you in \(\rm{}kWh\) $$ 1 \rm{kWh} = 10^3 (3600)\,\rm J \nonumber $$

Kirchhoff’s rules

Named after and first described by the German physicist Gustav Kirchhoff in 1845.

Junction rule

Charge conservation. At each junction, the current that flows in, must flow out. $$ \sum_i I_i = 0 \tag{Kirchhoff #1} \label{eq:Kirchhoff1} $$ The direction of the currents can be freely chosen.

Voltage rule

The sum of the voltage differences around any closed loop in a circuit must be zero. A loop in a circuit is any path which ends at the same point at which it starts. $$ \sum_i V_i = 0 \tag{Kirchhoff #2} \label{eq:Kirchhoff2} $$ E.g. as shown below \(-V_1+V_2+V_3=0\)

This is really the same as the work over a close line that we saw earlier $$ \oint \vec E\cdot d\vec l = 0 \nonumber $$



Given the circuit shown below. What are the currents through the resistors?

Apply Kirchhoff’s junction rule where the three resistors connect

Current’s \(I_1\) and \(I_2\) flow in, and \(I_3\) goes out. $$ \begin{align*} \sum_i I_i &= 0 \\ \Rightarrow I_1 + I_2 – I_3 &= 0 \\ \Rightarrow I_3 &= I_1 + I_2 \end{align*} $$

Substitute \(I_3\) in the schematic

Apply Kirchhoff’s voltage rule on the closed loop on the left $$ \begin{align*} \sum_i V_i &= 0 \\ \Rightarrow V_1+V_3-V_{b1} &= 0 & (V=IR) \\ \Rightarrow I_1R_1 + (I_1+I_2)R_3 – V_{bl} &= 0 \end{align*} $$

Take a closed loop on the right

Apply Kirchhoff’s voltage rule on the closed loop on the right $$ \begin{align*} \sum_i V_i &= 0 \\ \Rightarrow V_2+V_3+V_{b1} &= 0 & (V=IR) \\ \Rightarrow I_2R_2 + (I_1+I_2)R_3 + V_{b2} &= 0 \end{align*} $$

We now have two equations with two unknowns, \(I_1\) and \(I_2\), that we can solve $$ \begin{align*} I_1R_1 + (I_1+I_2)R_3 – V_{bl} &= 0 \\[0.3em] I_2R_2 + (I_1+I_2)R_3 + V_{b2} &= 0 \end{align*} $$

When substituting values, the currents may come out negative. That just means we assumed the wrong direction. The answer would still be correct.

Two (Lec.12)

Given the circuit given below, what are the currents and potential differences

To calculate the potentials differences, the parallel circuit of resistors \(R_1\ldots R_3\) can be replaced by \(R_p\) $$ \begin{align*} \frac{1}{R_p} &= \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_p} \\ \Rightarrow R_p &= \frac{1}{\frac{1}{1} + \frac{1}{2} + \frac{1}{3}} \approx 0.5455\,\rm\Omega \\ \end{align*} $$

Replacement schematic

The potential differences follow from Kirchhoff’s voltage rule, \(\sum_i V_i=0\) $$ \begin{align*} \mathcal{E} &= IR_i + IR_p + IR_4 = I(R_i+R_p+R_4) \\ \Rightarrow I &= \frac{\mathcal{E}}{R_i+R_p+R_4} \\ &= \frac{10}{0.1 + 0.5455 + 4} \approx 2.153\,\rm{A} \end{align*} $$

Back to the original schematic

This gives us potential difference \(V_i\), \(V_p\) and \(V_4\), and indirectly the currents through \(R_1\ldots R_3\) $$ \left\{ \begin{align*} V_i &= I R_i = 2.153\,(0.1) \approx 0.2143\,\rm V \\ V_4 &= I R_4 = 2.153\,(4) \approx 8.611\,\rm V \\ V_p &= I R_p = 2.153\,(0.5455) \approx 1.174\,\rm V \Rightarrow \end{align*} \\ \right. \Rightarrow \left\{ \begin{aligned} I_1 &= \frac{V_p}{R_1} = \frac{1.174}{1} \approx 1.174\,\rm A \nonumber \\ I_2 &= \frac{V_p}{R_2} = \frac{1.174}{2} \approx 0.587\,\rm A \nonumber \\ I_3 &= \frac{V_p}{R_3} = \frac{1.174}{3} \approx 0.391\,\rm A \nonumber \end{aligned} \right. $$

Embedded software developer
Passionately curious and stubbornly persistent. Enjoys to inspire and consult with others to exchange the poetry of logical ideas.

Leave a Reply

Your email address will not be published. Required fields are marked *


This site uses Akismet to reduce spam. Learn how your comment data is processed.