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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.
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Solution
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
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