JEE Mains · Physics · STD 12 -7. Alternating current
A \(110 \; V , 50 \; Hz , AC\) source is connected in the circuit (as shown in figure). The current through the resistance \(55 \; \Omega\), at resonance in the circuit, will be \(\dots \; A\).
- A \(0\)
- B \(1\)
- C \(2\)
- D \(3\)
Answer & Solution
Correct Answer
(A) \(0\)
Step-by-step Solution
Detailed explanation
At resonance \(I _{ L }= I _{ C }\) Alternatively, \(\frac{1}{Z}=\sqrt{\left(\frac{1}{X_{L}}-\frac{1}{X_{C}}\right)^{2}}\) At resonance, \(X _{ L }= X _{ C }\) and \(Z \rightarrow \infty\) \(\therefore Z_{\text {total circuit }} \rightarrow \infty\) i.e, \(I=0\) Ans. \(0\)
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