MHT CET · Physics · Alternating Current
In the circuit given below, the current through inductor is \(0.9 \mathrm{~A}\) and through the capacitor is \(0.6 \mathrm{~A}\). The current drawn from the a.c. source is
- A \(1.5 \mathrm{~A}\)
- B \(0.9 \mathrm{~A}\)
- C \(0.6 \mathrm{~A}\)
- D \(0.3 \mathrm{~A}\)
Answer & Solution
Correct Answer
(D) \(0.3 \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
As the currents in an inductor and capacitor are out of phase by \(180^{\circ}\), we can write Current through the capacitor \(\mathrm{I}_{\mathrm{C}}=0.9 \mathrm{~A}\) Current through the inductor \(\mathrm{I}_{\mathrm{L}}=-(0.6 \mathrm{~A})\)
\(\therefore \text {Total current drawn from the source}\) \(=\mathrm{I}_{\mathrm{C}}+\mathrm{I}_{\mathrm{L}}\)
\(=0.9-0.6\)
\(=0.3 \mathrm{~A}\)
\(\therefore \text {Total current drawn from the source}\) \(=\mathrm{I}_{\mathrm{C}}+\mathrm{I}_{\mathrm{L}}\)
\(=0.9-0.6\)
\(=0.3 \mathrm{~A}\)
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