An electric lamp designed for operation on 110 V C is connected to a 220 V AC supply, through a choke coil of inductance 2H, for proper operation. The angular frequency of the AC is $100 \sqrt{10} r a d / s$ . If a capacitor is to be used in the place of the choke coil, its capacitance must be

1 μ F

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2 μ F

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5 μ F

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10 μ F

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Solution

The correct option is C $5 μ F$
$V = \sqrt{V_{R}^{2} + V_{L}^{2}} = \sqrt{V_{R}^{2} + V_{C}^{2}}; V_{L} = V_{C}$
$L w = \frac{1}{w c}; C = \frac{1}{L w^{2}}$
$C = \frac{1}{2 \times 10^{4} \times 10} = 0.5 \times 10^{- 5}$
$C = 5 μ F$

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An electric lamp designed for operation on 110 V C is connected to a 220 V AC supply, through a choke coil of inductance 2H, for proper operation. The angular frequency of the AC is 100√10rad/s . If a capacitor is to be used in the place of the choke coil, its capacitance must be

The correct option is C 5μFV=√V2R+V2L=√V2R+V2C;VL=VC Lw=1wc;C=1Lw2 C=12×104×10=0.5×10−5 C=5μF

An electric lamp designed for operation on 110 V C is connected to a 220 V AC supply, through a choke coil of inductance 2H, for proper operation. The angular frequency of the AC is $100 \sqrt{10} r a d / s$ . If a capacitor is to be used in the place of the choke coil, its capacitance must be

The correct option is C $5 μ F$
$V = \sqrt{V_{R}^{2} + V_{L}^{2}} = \sqrt{V_{R}^{2} + V_{C}^{2}}; V_{L} = V_{C}$
$L w = \frac{1}{w c}; C = \frac{1}{L w^{2}}$
$C = \frac{1}{2 \times 10^{4} \times 10} = 0.5 \times 10^{- 5}$
$C = 5 μ F$