(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) It is clear that 1 A current flows in the circuit from B to A. 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
or 38 = 7 I₁ – 5 I₂ ...(i)
Similarly for the loop ABQA,
4I₂ + I₂ + 18 + 2 (I₁ + I₂) + (I₁ + I₂) + 7 = 0
or, 3 I₁ + 8 I₂ = – 25 ...(ii)
Solving equation (i) and (ii) for I₁ and I₂, we get
I₁ = 2.52 and I₂ = – 4.07 A
Hence, V_ab = – 5 x (4.07) + 18
= – 20.35 + 18
= – 2.35 V.