Question 288: AC fundamentals

Question

An a.c. e.m.f. of peak value $=230 \mathrm{~V}$ and frequency 50 Hz is connected to a circuit with $\mathrm{R}=11.5 \Omega, \mathrm{~L}=2.5 \mathrm{H}$ and a capacitor all in series. The value of capacitance is ' C ' for the current in the circuit to be maximum. The value of C and maximum current are respectively $\left(\pi^2=10\right)$.

Accepted Answer

Maximum current occurs at Resonance
$$ \begin{array}{ll} 	herefore & X_L=X_C \Rightarrow \omega L=\frac{1}{\omega C} \ & \omega^2=\frac{1}{L C} \Rightarrow C=\frac{1}{\omega^2 L} \ & \omega=100 \pi \Rightarrow \omega^2=(100 \pi)^2=10^4 .10=10^5 \ 	herefore & C=\frac{1}{10^5 	imes 2.5}=\frac{4}{10^6}=4 	imes 10^{-6} \mathrm{~F}=4 \mu \mathrm{~F} \end{array} $$
Maximum current at resonance
$$ \mathrm{I}_{\max }=\frac{\mathrm{V}_0}{\mathrm{R}}=\frac{230}{11.5}=20 \mathrm{~A} $$
$	herefore \quad \mathrm{C}=4 \mu \mathrm{~F}$ and maximum current $\mathrm{I}_{\max }=20 \mathrm{~A}$

Answer

$2 \mu \mathrm{~F}, 10 \mathrm{~A}$

Answer

$4 \mu \mathrm{~F}, 20 \mathrm{~A}$

Answer

$6 \mu \mathrm{~F}, 10 \mathrm{~A}$

Answer

$8 \mu \mathrm{~F}, 20 \mathrm{~A}$

AC fundamentals

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