Question 281: AC fundamentals

Question

An a.c. e.m.f. of peak value 230 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 ( $\pi^2=10$ )

Accepted Answer

Step 1: Resonance condition
At resonance:
$ \omega L = \frac{1}{\omega C} $
where $\omega = 2\pi f$.
So:
$ C = \frac{1}{\omega^2 L} $
Step 2: Angular frequency
$ f = 50 \,	ext{Hz}, \quad \omega = 2\pi f = 100\pi \, 	ext{rad/s} $
Step 3: Capacitance
$ C = \frac{1}{(100\pi)^2 \cdot 2.5} $
Given: $\pi^2 = 10 \Rightarrow (100\pi)^2 = 10{,}000 \pi^2 = 100{,}000$.
So:
$ C = \frac{1}{2.5 	imes 100{,}000} = \frac{1}{250{,}000} \, 	ext{F} $
$ C = 4 	imes 10^{-6} \,	ext{F} = 4 \,\mu	ext{F} $
Step 4: Maximum Current at resonance
At resonance, impedance = $R$.
So peak current:
$ I_0 = \frac{V_0}{R} = \frac{230}{11.5} = 20 \,	ext{A} $
Final Answer:
$ C = 4 \,\mu	ext{F}, \quad I_{\max} = 20 \,	ext{A} $
Correct option: A ( $4 \mu 	ext{F}, 20 	ext{A}$ )
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Answer

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

Answer

$5 \mu \mathrm{~F}, 10 \mathrm{~A}$

Answer

$2 \mu \mathrm{~F}, 20 \mathrm{~A}$

Answer

$8 \mu \mathrm{~F}, 12 \mathrm{~A}$

AC fundamentals

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