First slide

Electrical power

Question

A torch bulb rated as 4.5 W, 1.5 V is connected as shown in fig. The e.m.f. of the cell, needed to
make the bulb glow at full intensity is

Moderate

A

4.5 V
B

1.5 V
C

2.67 V
D

13.5 V

Solution

Resistance of bulb $R_{b} = \frac{(1 \cdot 5)^{2}}{4 \cdot 5} = 0 \cdot 5 Ω$
Current drawn from battery $= \frac{E}{2 \cdot 67 + 0 \cdot 33} = \frac{E}{3}$
Share of bulb $\frac{2}{3} \times \frac{E}{3} = \frac{2 E}{9}$
$∴ {(\frac{2 E}{9})}^{2} \times 0 \cdot 5 = 4 \cdot 5$
or $E = 13 \cdot 5 V$

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