KCET · Physics · Alternating Current
A series LCR circuit containing an AC source of 100 V has an inductor and a capacitor of reactances \(24 \Omega\) and \(16 \Omega\) respectively. If a resistance of \(6 \Omega\) is connected in series, then the potential difference across the series combination of inductor and capacitor only is
- A 80 V
- B 400 V
- C 8 V
- D 40 V
Answer & Solution
Correct Answer
(A) 80 V
Step-by-step Solution
Detailed explanation
\(\mathrm{X}_{\mathrm{C}}=24 \Omega \)
\( \mathrm{X}_{\mathrm{L}}=16 \Omega \)
\( \mathrm{R}=6 \Omega \)
\( \mathrm{Z}=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}\right)^2} \)
\( =\sqrt{6^2+8^2} \)
\( =\sqrt{100}=10 \Omega\)
\(\mathrm{i}=\frac{\mathrm{V}}{\mathrm{Z}}=\frac{100}{10}=10 \mathrm{~A}\)
Net voltage across capacitor and inductor
\(=\mathrm{i}\left(\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}\right) \)
\(=10(24-16) \)
\(=10(8) \)
\(=80 \mathrm{~V}\)
\( \mathrm{X}_{\mathrm{L}}=16 \Omega \)
\( \mathrm{R}=6 \Omega \)
\( \mathrm{Z}=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}\right)^2} \)
\( =\sqrt{6^2+8^2} \)
\( =\sqrt{100}=10 \Omega\)
\(\mathrm{i}=\frac{\mathrm{V}}{\mathrm{Z}}=\frac{100}{10}=10 \mathrm{~A}\)
Net voltage across capacitor and inductor
\(=\mathrm{i}\left(\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}\right) \)
\(=10(24-16) \)
\(=10(8) \)
\(=80 \mathrm{~V}\)
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