Question

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

1. Current in circuit 

2. The terminal voltage of the cell 

3. P.D. across 6Ω resistor. 

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

Answer 1

 Given, Emf (E) = 12 V; r₁ = 2Ω ; RA = 4Ω ; RB = 6Ω

1. The current in the circuit is 

1 = E/R_total = E/R₁ + R_A + R_B 

I = 12/2 + 4 + 6 = IA 

2. The terminal voltage of the cell is 

Terminal Voltage = Emf- Ir, 

Terminal Voltage= 12 – (1 × 2) = 12 – 2 = 10V 

3. The potential difference across the 6Ω resistor is V_B = IR_B 

∴ VB = 1 x 6 = 6V 

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

E = I²Rt 

E = 1² × 4 × 60 = 240J