Question

A cell of e.m.f. 1.5 V and internal resistance 1.0 W is connected to two resistors of 4.0 W and 20.0 in series as shown in the figure: Calculate the:

(i) Current in the circuit.
(ii) Potential difference across the 4.0 ohm resistor.
(iii) Voltage drop when the current is flowing.
(iv) Potential difference across the cell.

Answer 1

E.m.f. of cell = 1.5 volt, Internal resistance = 1.0 ohm

External resistance = r₁ + r₂ (in series) 4 + 20 \ 24 Ω

current (I) = `"E"/("r" + "R")` or I = `1.5/(1 + 24) = 1.5/25`

= 0.06 Amp.

(ii) ∴ P.D. across 4 Ω resistor = r₁I = 4 × 0.06 = 0.24 volt

(iii) Voltage drop = Ir = 0.06 × 1 = 0.06 volt

(iv) ∴ Potential across cell = RI = 24 × 0.06

= 1.44 volt