Question

In the given figure, a steady current I flows through the circuit when points A and C are connected by a wire of negligible resistance. Find the potential difference between points B and C.

Answer 1

Given: E₁ = 6 V, internal resistance r₁ = 1 Ω

E₂ = 4 V, internal resistance r₂ = 3 Ω

Net current in the loop.

E_net = 6 − 4

= 2 V

R_total = 1 + 3 Ω

= 4 Ω

I = `(E_"net")/(R_"total")`

= `2/4`

= 0.5 A

Potential difference between B and C

Since current enters the positive terminal, the cell is being charged.

V_BC = E − Ir

= 4 − (0.5 × 3)

= 4 − 1.5

= 2.5 V