In the circuit given below,if the bulb is to | vedclass

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(A) The bulb is rated at $V_b = 1.5 	ext{ V}$ and $P = 0.45 	ext{ W}$.The resistance of the bulb is $R_b = \frac{V_b^2}{P} = \frac{(1.5)^2}{0.45} =

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$1.25$

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(A) The bulb is rated at $V_b = 1.5 	ext{ V}$ and $P = 0.45 	ext{ W}$.The resistance of the bulb is $R_b = \frac{V_b^2}{P} = \frac{(1.5)^2}{0.45} = \frac{2.25}{0.45} = 5 \Omega$.For the bulb to glow with maximum intensity,it must operate at its rated voltage of $1.5 	ext{ V}$.The current through the bulb is $i_1 = \frac{P}{V_b} = \frac{0.45}{1.5} = 0.3 	ext{ A}$.The voltage across the $3 \Omega$ resistor is $V_{3\Omega} = 6 	ext{ V} - 1.5 	ext{ V} = 4.5 	ext{ V}$.The total current in the circuit is $i = \frac{V_{3\Omega}}{3 \Omega} = \frac{4.5}{3} = 1.5 	ext{ A}$.The current through the resistor $R$ is $i - i_1 = 1.5 	ext{ A} - 0.3 	ext{ A} = 1.2 	ext{ A}$.Since the resistor $R$ is in parallel with the bulb,the voltage across $R$ is $1.5 	ext{ V}$.Therefore,$R = \frac{1.5 	ext{ V}}{1.2 	ext{ A}} = 1.25 \Omega$.

In the circuit given below,if the bulb is to glow with maximum intensity,the value of $R$ is (neglect internal resistance of the cell) (in $Omega$)

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