MHT CET · Physics · Alternating Current
A series LCR circuit containing a resistance 'R' has angular frequency ' \(\omega\) '. At resonance the voltage across resistance and inductor are ' \(\mathrm{V}_{\mathrm{R}}\) ' and ' \(\mathrm{V}_{\mathrm{L}}\) ' respectively, then value of inductance 'L' will be
- A \(\frac{V_R R}{V_L \omega}\)
- B \(\frac{V_L}{V_R R \omega}\)
- C \(\frac{V_R \omega}{V_L R}\)
- D \(\frac{\mathrm{V}_{\mathrm{L}} \mathrm{R}}{\mathrm{V}_{\mathrm{R}}(\omega)}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{V}_{\mathrm{L}} \mathrm{R}}{\mathrm{V}_{\mathrm{R}}(\omega)}\)
Step-by-step Solution
Detailed explanation
At resonance,
\(Z=R\)
Voltage across resistance,
\(V_R=I \times Z=I \times R ...(i)\)
Voltage across inductor,
\(\mathrm{V}_{\mathrm{L}}=\mathrm{I} \times \mathrm{X}_{\mathrm{L}}=\mathrm{I} \times \omega \mathrm{L}...(ii)\)
Dividing (ii) by (i),
\(\frac{\mathrm{V}_{\mathrm{L}}}{\mathrm{~V}_{\mathrm{R}}}=\frac{\omega \mathrm{L}}{\mathrm{R}} \Rightarrow \mathrm{~L}=\frac{\mathrm{V}_{\mathrm{L}} \mathrm{R}}{\mathrm{~V}_{\mathrm{R}} \omega}\)
\(Z=R\)
Voltage across resistance,
\(V_R=I \times Z=I \times R ...(i)\)
Voltage across inductor,
\(\mathrm{V}_{\mathrm{L}}=\mathrm{I} \times \mathrm{X}_{\mathrm{L}}=\mathrm{I} \times \omega \mathrm{L}...(ii)\)
Dividing (ii) by (i),
\(\frac{\mathrm{V}_{\mathrm{L}}}{\mathrm{~V}_{\mathrm{R}}}=\frac{\omega \mathrm{L}}{\mathrm{R}} \Rightarrow \mathrm{~L}=\frac{\mathrm{V}_{\mathrm{L}} \mathrm{R}}{\mathrm{~V}_{\mathrm{R}} \omega}\)
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