(D) For the inductor coil at $50 \, Hz$:$X_L = \frac{V}{I} = \frac{100}{8} = 12.5 \, \Omega$.For the resistor:$R = \frac{V}{I} = \frac{100}{10} = 10 \

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(D) For the inductor coil at $50 \, Hz$:$X_L = \frac{V}{I} = \frac{100}{8} = 12.5 \, \Omega$.For the resistor:$R = \frac{V}{I} = \frac{100}{10} = 10 \, \Omega$.At the new frequency $f' = 40 \, Hz$, the new inductive reactance $X_L'$ is:$X_L' = X_L \times \frac{f'}{f} = 12.5 \times \frac{40}{50} = 10 \, \Omega$.When connected in series, the impedance $Z$ is:$Z = \sqrt{R^2 + (X_L')^2} = \sqrt{10^2 + 10^2} = 10\sqrt{2} \, \Omega$.The current $I$ in the series circuit is:$I = \frac{V}{Z} = \frac{100}{10\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2} \, A$.

Class 12 Physics Alternating Current Inductance, Capacitance and Resistance in Series and Parallel

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An inductor coil takes a current of $8 \, A$ when connected to a $100 \, V$ and $50 \, Hz$ $AC$ source. $A$ pure resistor under the same condition takes a current of $10 \, A$. If the inductor coil and resistor are connected in series to a $100 \, V$ and $40 \, Hz$ $AC$ supply, then the current in the series combination of the above resistor and inductor is:

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A
$\frac{10}{\sqrt{3}} \, A$
B
$\frac{5}{\sqrt{2}} \, A$
C
$10 \sqrt{2} \, A$
D
$5 \sqrt{2} \, A$

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Alternating Current

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Inductance, Capacitance and Resistance in Series and Parallel

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In a series $LCR$ circuit,if the current leads the source voltage,then

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In a series $RLC$ circuit,the $r.m.s.$ voltages across the resistor and the inductor are respectively $400 \,V$ and $700 \,V$. If the equation for the applied voltage is $\varepsilon = 500 \sqrt{2} \sin \omega t$,then the peak voltage across the capacitor is ........... $V$.

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