This law deals with e.m.f.s and voltages in a circuit. We will start by considering a simple circuit that contains a cell and two resistors of resistances ${R_1}$ and ${R_2}$ (Figure 9.8). Since this is a simple series circuit, the current I must be the same all the way around, and we need not concern ourselves further with Kirchhoff’s first law. For this circuit, we can write the following equation:

$E = I{R_1} + I{R_2}$
e.m.f. of battery = sum of p.d.s across the resistors

You should not find these equations surprising. However, you may not realise that they are a consequence of applying Kirchhoff’s second law to the circuit. This law states that the sum of the e.m.f.s around any loop in a circuit is equal to the sum of the p.d.s around the loop.

WORKED EXAMPLE

1) Use Kirchhoff’s laws to find the current in the circuit in Figure 9.9.
This is a series circuit so the current is the same all the way round the circuit.

Step 1: We calculate the sum of the e.m.f.s:
$sum\,of\,e.m.f.s = 6.0V - 2.0V = 4.0V$
The batteries are connected in opposite directions so we must consider one of the e.m.f.s as negative.
Step 2: We calculate the sum of the p.d.s.
$sum\,of\,p.d.s = (I \times 10) + (I \times 30) = 40I$
Step 3: We equate these:
$4.0 - 40I$
and so $I = 0.1A$
No doubt, you could have solved this problem without formally applying Kirchhoff’s second law, but you will find that in more complex problems the use of these laws will help you to avoid errors.
You will see later that Kirchhoff’s second law is an expression of the conservation of energy.
We shall look at another example of how this law can be applied, and then look at how it can be applied in general.

Question

 

5) Use Kirchhoff’s second law to deduce the p.d. across the resistor of resistance R in the circuit shown in Figure 9.10, and hence find the value of R. (Assume the battery of e.m.f. $10 V$ has negligible internal resistance.)

An equation for Kirchhoff’s second law

In a similar manner to the formal statement of the first law, the second law can be written as an equation:

$\Sigma E = \Sigma V$

where $\Sigma E$ is the sum of the e.m.f.s and $\Sigma V$ is the sum of the potential differences.