(a) Find the emf E₁ and E₂ in the circuit of the following diagram and the potential difference between the points a and b.
(b) If in the above circuit, the polarity of the battery E₁, be reversed, what will be the potential difference between a and b?

(a)From the fig. below it is evident that 1 Ampere of current flows in the circuit from B to A.



On applying Kirchhoff’s law to the loop PAQBP,

20 – E₂ = (12 x 1) + (1 x 2) + (2 x 2) = 18

Hence, E₂ = 2 V .

Thus the potential difference between the points A and B is,
V_AB = 18 – 1 – 4 = 13 V.

(b) On reversing the polarity of the battery E₁, the current distributions will be changed.
Let the currents be I₁ and I₂ as shown in the following figure.

Applying Kirchhoff’s law for the loop PABP,

20 + E₁ = (6 + 1) I₁ – (4 + 1) I₂
$\Rightarrow$38 = 7 I₁ – 5 I₂                      ...(i)

Similarly for the loop ABQA,

4I₂ + I₂ + 18 + 2 (I₁ + I₂) + (I₁ + I₂) + 7 = 0
$\Rightarrow$3 I₁ + 8 I₂ = – 25                   ...(ii)

Solving equation (i) and (ii) for I₁ and I₂, we get 

I₁ = 2.52 A and I₂ = – 4.07 A

Hence, 
V_ab = – 5 x (4.07) + 18
      = – 20.35 + 18
      = – 2.35 V.