JEE MAIN - Physics (2005 - No. 15)
The self inductance of the motor of an electric fan is $$10$$ $$H$$. In order to impart maximum power at $$50$$ $$Hz$$, it should be connected to a capacitance of
$$8\mu F$$
$$4\mu F$$
$$2\mu F$$
$$1\mu F$$
Explanation
For maximum power, $${X_L} = X{}_C,$$ which yields
$$C = {1 \over {{{\left( {2\pi n} \right)}^2}L}} = {1 \over {4{\pi ^2} \times 50 \times 50 \times 10}}$$
$$\therefore$$ $$C = 0.1 \times {10^{ - 5}}F = 1\mu F$$
$$C = {1 \over {{{\left( {2\pi n} \right)}^2}L}} = {1 \over {4{\pi ^2} \times 50 \times 50 \times 10}}$$
$$\therefore$$ $$C = 0.1 \times {10^{ - 5}}F = 1\mu F$$
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