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The applied potential difference in the circuit shown is V=10sin100t where V is in volt and t is in second. If the power factor of the circuit is 1/√2, then the value of capacitance C of the circuit nearly is:

a) 111 μF b) 333 μF c) 222 μF d) 444 μF Answer: a) 111 μF Solution: \(\begin{array}{l}cos\phi =\frac{X_{L}-X_{C}}{\sqrt{R^{2}+(X_{L}-X_{C})^{2}}}=\frac{100\times 0.1-\frac{1}{0.1C}}{\sqrt{20^{2}+(100\times 0.1-\frac{1}{0.1C})^{2}}}\end{array} \)   \(\begin{array}{l}\frac{1}{\sqrt{2}} =\frac{10-\frac{10}{C}}{\sqrt{400+(10-\frac{10}{C})^{2}}}\end{array} \)... View Article