MHT CET · Physics · Alternating Current
An inductor of inductance \(2 \mu \mathrm{H}\) is connected in series with a resistance, a variable capacitor and an a.c. source of frequency 5 kHz . The value of capacitance for which maximum current is drawn into the circuit is \(\frac{1}{\mathrm{x}} \mathrm{F}\), where the value of ' \(x\) ' is (Take \(\pi^2=10\) )
- A 500
- B 1000
- C 2000
- D 4000
Answer & Solution
Correct Answer
(C) 2000
Step-by-step Solution
Detailed explanation
For maximum current to be drawn the circuit should be in resonance,
\(\therefore \mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}} \quad \Rightarrow 2 \pi \mathrm{fL}=\frac{1}{2 \pi \mathrm{fC}}\)
\(\therefore \mathrm{C}=\frac{1}{4 \pi^2 \mathrm{f}^2 \mathrm{~L}}=\frac{1}{4 \times 10 \times 25 \times 10^6 \times 2 \times 10^{-6}}\)
\(\therefore \mathrm{C}=\frac{1}{2000} \mathrm{~F} \quad \Rightarrow \mathrm{x}=2000\)
\(\therefore \mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}} \quad \Rightarrow 2 \pi \mathrm{fL}=\frac{1}{2 \pi \mathrm{fC}}\)
\(\therefore \mathrm{C}=\frac{1}{4 \pi^2 \mathrm{f}^2 \mathrm{~L}}=\frac{1}{4 \times 10 \times 25 \times 10^6 \times 2 \times 10^{-6}}\)
\(\therefore \mathrm{C}=\frac{1}{2000} \mathrm{~F} \quad \Rightarrow \mathrm{x}=2000\)
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