MHT CET · Physics · Alternating Current
Resistor of \(2 \Omega\), inductor of \(100 \mu \mathrm{H}\) and capacitor of \(400 \mathrm{pF}\) are connected in series across an a.c. source of \(e_{r m s}=0.1\) volt. At resonance, voltage drop across inductor is
- A 20V
- B 25 V
- C 2.5 V
- D 250 V
Answer & Solution
Correct Answer
(B) 25 V
Step-by-step Solution
Detailed explanation
We know, at resonance \(X_C=X_L\)
\(\Rightarrow \omega L=\frac{1}{\omega C} \Rightarrow \omega=\left(\frac{1}{\sqrt{L C}}\right)\)
The potential drop across inductor is
\(V_L=i X_L=i \sqrt{\frac{L}{C}}\)
\(\because\) LCR circuit in series and at resonance
\(i=\frac{V}{R}=\frac{0.1}{2} \mathrm{~A}=0.05 \mathrm{~A}\)
Now, \(V_L=0.05 \sqrt{\left(\frac{100 \times 10^{-6}}{400 \times 10^{-12}}\right)} \mathrm{V}=2.5 \times 10^{-2} \times 10^3\)\(\mathrm{~V}=25 \mathrm{~V}\)
\(\Rightarrow \omega L=\frac{1}{\omega C} \Rightarrow \omega=\left(\frac{1}{\sqrt{L C}}\right)\)
The potential drop across inductor is
\(V_L=i X_L=i \sqrt{\frac{L}{C}}\)
\(\because\) LCR circuit in series and at resonance
\(i=\frac{V}{R}=\frac{0.1}{2} \mathrm{~A}=0.05 \mathrm{~A}\)
Now, \(V_L=0.05 \sqrt{\left(\frac{100 \times 10^{-6}}{400 \times 10^{-12}}\right)} \mathrm{V}=2.5 \times 10^{-2} \times 10^3\)\(\mathrm{~V}=25 \mathrm{~V}\)
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