A bulb with a rating of 4.5 , W, 1.5 , V is c | vedclass

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(D) The current through the bulb is $I_{bulb} = \frac{P}{V} = \frac{4.5 \, W}{1.5 \, V} = 3 \, A$.The current through the $1 \, \Omega$ resistor is $I

Vedclass · Accepted Answer

$13.5$

Vedclass · Answer

(D) The current through the bulb is $I_{bulb} = \frac{P}{V} = \frac{4.5 \, W}{1.5 \, V} = 3 \, A$.The current through the $1 \, \Omega$ resistor is $I_{resistor} = \frac{V}{R} = \frac{1.5 \, V}{1 \, \Omega} = 1.5 \, A$.The total current flowing from the cell is $I_{total} = I_{bulb} + I_{resistor} = 3 \, A + 1.5 \, A = 4.5 \, A$.The $e.m.f.$ of the cell is given by $E = V + I_{total} \cdot r$,where $V = 1.5 \, V$ and $r = 2.67 \, \Omega$.Substituting the values: $E = 1.5 + (4.5 	imes 2.67) = 1.5 + 12.015 \approx 13.5 \, V$.

$A$ bulb with a rating of $4.5 \, W, 1.5 \, V$ is connected as shown in the figure. To make the bulb glow at full brightness,the $e.m.f.$ of the cell should be ................ $V$.

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