Current Electricity

JEE Main level, numerical answer, medium

145 of 380

Question 145 of 380

An ideal cell of emf 10V10V is connected in circuit shown in figure. Each resistance is 2Ω2\Omega. The potential difference (in VV) across the capacitor when it is fully charged is _____________

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Answer: 8


i=1043+2=10×310=3Ampi=\dfrac{10}{\dfrac{4}{3}+2}=\dfrac{10\times 3}{10}=3Amp
{i}_{1}=2A&{i}_{2}=1A
VAB=1×2+3×2=8V{V}_{AB}=1\times 2+3\times 2=8V

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