Question

A battery of emf 12 V and internal resistance 2 Ω is connected with two resistors A and B of resistance 4 Ω and 6 Ω respectively joined in series.

Find:

1) Current in the circuit

2) The terminal voltage of the cell

3) The potential difference across 6Ω Resistor

4) Electrical energy spent per minute in the 4Ω resistor.

Answer 1

E = 12 V; r_i = 2 Ω; R_A = 4 Ω; R_B = 6 Ω

1) The current in the circuit is

`I = E/R_"total" = E/(r_i + r_A + r_B)`

`:. I = 12/(2+4+6) = 1 A`

2) The terminal voltage of the cell is

V = E - Ir_i

∴ V = 12 - (1 x 2) = 12 - 2 = 10 V

3) The potential difference across the 6 Ω is

V_B = IR_B

∴ V_B = 1 x 6 = 6V

4) The electrical energy spent per minute (=60 s) is

E = I²Rt

E = 1² X 4 X 60 = 240J