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An ideal cell of emf 10V is connected in circuit shown in figure. Each resistance is 2\Omega. The potential difference (in V) across the capacitor when it is fully charged is .........
Option: 1 2
Option: 2 4
Option: 3 6
Option: 4 8

Answers (1)

best_answer

when the capacitor is fully charged it behaves as an open circuit  (voltage drop with no current)

So the new circuit diagram is given below.

So

\begin{array}{l} \mathrm{i}=\frac{10}{\frac{4}{3}+2}=\frac{10 \times 3}{10}=3 \mathrm{Amp} \\ \Rightarrow \mathrm{i}_{1}=2 \mathrm{~A} \ \& \ \mathrm{i}_{2}=1 \mathrm{~A} \\ \Rightarrow \mathrm{~V}_{\mathrm{AB}}=1 \times 2+3 \times 2=8 \mathrm{~V} \end{array}

 

Posted by

avinash.dongre

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