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  4. An ideal inductor takes a current of 10 A when connected to a 125 V , 50 Hz AC supply. A pure resistor across the same source takes 12.5 A. if the two are connected in series across a 100√2 V 40Hz supply , the current through the circuit will be

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Q. An ideal inductor takes a current of 10 A when connected to a 125 V , 50 Hz AC supply. A pure resistor across the same source takes 12.5 A. if the two are connected in series across a $100\sqrt2 V$ 40Hz supply , the current through the circuit will be

Alternating Current

Solution:

For 50 Hz and 125 V supply
$X_L=\omega \,L =\frac{V}{i_L}\Rightarrow \,L=\frac{1}{8\pi},R=\frac{V}{i_R}=10 \Omega$
For 40 Hz, 100 $\sqrt{2\,V}$ supply
$i=\frac{V}{\sqrt{R^2+X^{2_L}}}=\frac{V}{\sqrt{R^2+4\pi^2\,f^2\,L^2}}$