KCET · Physics · Alternating Current
In a series L.C.R circuit, the potential drop across \( L, C \) and \( R \) respectively are \( 40 \mathrm{~V}, 120 \mathrm{~V} \) and \( 60 \mathrm{~V} \). Then the source voltage is
- A \( 220 \mathrm{~V} \)
- B \( 160 \mathrm{~V} \)
- C \( 180 \mathrm{~V} \)
- D \( 100 \mathrm{~V} \)
Answer & Solution
Correct Answer
(D) \( 100 \mathrm{~V} \)
Step-by-step Solution
Detailed explanation
Given, voltage drop across \( \mathrm{L}, V_{L}=40 \mathrm{~V} ; \) voltage dropacross \( \mathrm{C}, V_{c}=120 \mathrm{~V} ; \) voltage drop across \( R, V_{R}=60 \mathrm{~V} \)
Then source voltage \( =\sqrt{{V_{R}}^{2}+\left(V_{L}-V_{C}\right)^{2}} \)
\( =\sqrt{60^{2}+(120-40)^{2}} \)
\( =\sqrt{3600+6400} \)
\( =\sqrt{10000} \)
\( =100 \mathrm{~V} \)
Therefore, source voltage \( =100 \mathrm{~V} \)
Then source voltage \( =\sqrt{{V_{R}}^{2}+\left(V_{L}-V_{C}\right)^{2}} \)
\( =\sqrt{60^{2}+(120-40)^{2}} \)
\( =\sqrt{3600+6400} \)
\( =\sqrt{10000} \)
\( =100 \mathrm{~V} \)
Therefore, source voltage \( =100 \mathrm{~V} \)
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