Question

A cell of e.m.f 2.0 V and internal resistance 1Ω is connected to the resistors of 3Ω and 6Ω in series. Calculate:
(i) the current drawn from the cell,
(ii) the p.d. across each resistor,
(iii) the terminal voltage of the cell and
(iv) the voltage drop.

Answer 1

Total resistance of the circuit R = 1 + 3 + 6 = 10 Ω

(i) The current drawn from the cell = `"E"/"R" = 2.0/10 = 0.2` A

(ii) The p.d. across 3 Ω resistor V₁ = IR₁ = 0.2 × 3 = 0.6V 

The p.d. across 6 Ω resistor V₂ = IR₂ = 0.2 × 6 = 1.2 V 

(iii) The terminal voltage of the cell V = V₁ + V₂

= 0.6 + 1.2 = 1.8 V

(iv) The voltage drop = E - V = 2.0 - 1.8 = 0.2 V