A torch bulb rated as 4.5, W, 1.5, V is conne | vedclass

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(D) The resistance of the bulb is given by $R_b = \frac{V^2}{P} = \frac{(1.5)^2}{4.5} = \frac{2.25}{4.5} = 0.5\, \Omega$.The bulb is in parallel with

Vedclass · Accepted Answer

$13.5$

Vedclass · Answer

(D) The resistance of the bulb is given by $R_b = \frac{V^2}{P} = \frac{(1.5)^2}{4.5} = \frac{2.25}{4.5} = 0.5\, \Omega$.The bulb is in parallel with a resistor of $1\, \Omega$. The equivalent resistance of this parallel combination is $R_p = \frac{R_b 	imes 1}{R_b + 1} = \frac{0.5 	imes 1}{0.5 + 1} = \frac{0.5}{1.5} = \frac{1}{3}\, \Omega$.The current flowing through the bulb is $I_b = \frac{P}{V} = \frac{4.5}{1.5} = 3\, A$.The current flowing through the $1\, \Omega$ resistor is $I_r = \frac{V}{R} = \frac{1.5}{1} = 1.5\, A$.The total current drawn from the cell is $I = I_b + I_r = 3 + 1.5 = 4.5\, A$.The internal resistance of the cell is $r = 2.67\, \Omega$ (approximately $\frac{8}{3}\, \Omega$).The $e.m.f.$ of the cell is $E = V + Ir = 1.5 + (4.5 	imes \frac{8}{3}) = 1.5 + (1.5 	imes 8) = 1.5 + 12 = 13.5\, V$.

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

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