JEE Mains · Physics · STD 12 -7. Alternating current
A circuit element \(X\) when connected to an a.c. supply of peak voltage \(100\,V\) gives a peak current of \(5\,A\) which is in phase with the voltage. A second element \(Y\) when connected to the same a.c. supply also gives the same value of peak current which lags behind the voltage by \(\frac{\pi}{2}\). If \(X\) and \(Y\) are connected in series to the same supply, what will be the rms value of the current in ampere?
- A \(\frac{10}{\sqrt{2}}\)
- B \(\frac{5}{\sqrt{2}}\)
- C \(5 \sqrt{2}\)
- D \(\frac{5}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{5}{2}\)
Step-by-step Solution
Detailed explanation
Element \(X\) should be resistive with \(R =20 \Omega\) Element \(Y\) should be inductive with \(X _{ L }=20 \Omega\) When \(X\) and \(Y\) are connector in series \(Z =\sqrt{ X _{ L }^{2}+ R ^{2}}=20 \sqrt{2}\)…
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