KCET · Physics · Alternating Current
In \(R-L-C\) series circuit, the potential differences across each element is \(20 \mathrm{~V}\). Now the value of the resistance alone is doubled, then P.D. across \(R, L\) and \(C\) respectively.
- A \(20 \mathrm{~V}, 10 \mathrm{~V}, 10 \mathrm{~V}\)
- B \(20 \mathrm{~V}, 20 \mathrm{~V}, 20 \mathrm{~V}\)
- C \(20 \mathrm{~V}, 40 \mathrm{~V}, 40 \mathrm{~V}\)
- D \(10 \mathrm{~V}, 20 \mathrm{~V}, 20 \mathrm{~V}\)
Answer & Solution
Correct Answer
(A) \(20 \mathrm{~V}, 10 \mathrm{~V}, 10 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
Given, the potential difference across each element is same i.e., \(20 \mathrm{~V}\), so, the circuit is at resonance and we have
\(V=V_{R}=20 \mathrm{~V}\)
If the value of \(R\) is doubled, then value of / reduced to half \(\therefore\)
\(V_{L}=V_{C}=\frac{I X_{L}}{2}=\frac{I X_{C}}{2}=\frac{20}{2}=10 \mathrm{~V}\)
\((\because\) circuit is at resonance) and \(\quad V_{R}=V=20 \mathrm{~V}\)
\(V=V_{R}=20 \mathrm{~V}\)
If the value of \(R\) is doubled, then value of / reduced to half \(\therefore\)
\(V_{L}=V_{C}=\frac{I X_{L}}{2}=\frac{I X_{C}}{2}=\frac{20}{2}=10 \mathrm{~V}\)
\((\because\) circuit is at resonance) and \(\quad V_{R}=V=20 \mathrm{~V}\)
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