JEE Mains · Physics · STD 12 -7. Alternating current
A capacitor of capacitance \(150.0\,\mu F\) is connected to an alternating source of \(emf\) given by \(E =36\) \(\sin (120 \pi t ) V\). The maximum value of current in the circuit is approximately equal to \(......\,A\)
- A \(2\)
- B \(\frac{1}{\sqrt{2}}\)
- C \(\sqrt{2}\)
- D \(2 \sqrt{2}\)
Answer & Solution
Correct Answer
(A) \(2\)
Step-by-step Solution
Detailed explanation
\(I_0=\frac{E_0}{x_c}=\frac{E_0}{\frac{1}{\omega_c}}=E_0 \omega_c\) \(\Rightarrow I_0=36 \times 120 \pi \times 150 \times 10^{-6}\) \(\Rightarrow I_0=2.03\) \(\simeq 2\,A\)
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