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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.
Solution
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
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