Question

Four resistors of resistance 0.5 Ω, 1.5 Ω, 4 Ω and 6 Ω are connected in series to a battery of e.m.f. 6 V and negligible internal resistance. Calculate (a) current drawn from the cell (b) p.d. at the ends of each resistor.

Answer 1

Total resistance of 0.5Ω, 1.5Ω, 4Ω and 6Ω in series,

\[ R = 0.5 + 1.5 + 4 + 6 \] \[ = 12\Omega \]

(a) \[ \therefore \text{Current drawn from cell (I)} = \frac{V}{R} \] \[ = \frac{6}{12} \] \[ = 0.5 \text{A}\]

(b) P.D. across 0.5Ω resistor, \[ V_1 = I.R \] \[ = 0.5 \times 0.5 \] \[ = 0.25 \text{V} \]

P.D. across 1.5Ω resistor, \[ V_2 = I.R \] \[ = 0.5 \times 1.5 \] \[ = 0.75 \text{V} \]

P.D. across 4Ω resistor, \[ V_3 = I.R \] \[ = 0.5 \times 4 \] \[ = 2 \text{V} \]

P.D. across 6Ω resistor, \[ V_4 = I.R \] \[ = 0.5 \times 6 \] \[ = 3 \text{V} \]