MHT CET · Physics · Alternating Current
In the given circuit diagram, in the steady state the current through the battery and the charge on the capacitor respectively are
- A 2 A and \(3 \mu \mathrm{C}\)
- B \(\frac{6}{11} \mathrm{~A}\) and \(\frac{12}{7} \mu \mathrm{C}\)
- C 11 A and \(3 \mu \mathrm{C}\)
- D zero ampere and \(3 \mu \mathrm{C}\)
Answer & Solution
Correct Answer
(C) 11 A and \(3 \mu \mathrm{C}\)
Step-by-step Solution
Detailed explanation
In steady state, current through capacitor is zero, hence current through \(4 \Omega\) resistor becomes zero.
\(\begin{array}{ll}
\therefore & \frac{1}{R_{e q}}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}=\frac{6+3+2}{6}=\frac{11}{6} \\
\therefore & I=\frac{V}{R_{e q}}=\frac{6 \times 11}{6}=11 \mathrm{~A}
\end{array}\)
\(\therefore \quad\) Charge on capacitor is, \(\mathrm{Q}_{\mathrm{C}}=\mathrm{CV}=0.5 \times 6=3 \mu \mathrm{C}\)
\(\begin{array}{ll}
\therefore & \frac{1}{R_{e q}}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}=\frac{6+3+2}{6}=\frac{11}{6} \\
\therefore & I=\frac{V}{R_{e q}}=\frac{6 \times 11}{6}=11 \mathrm{~A}
\end{array}\)
\(\therefore \quad\) Charge on capacitor is, \(\mathrm{Q}_{\mathrm{C}}=\mathrm{CV}=0.5 \times 6=3 \mu \mathrm{C}\)
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