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JEE Mains · Physics · STD 12 - Current electricity
  • JEE Mains 2014
  • Hard
  • MCQ

The circuit shown here has two batteries of \(8.0\,V\) and \(16.0\,V\) and three resistors \(3\,\Omega ,\,9\,\Omega \) and \(9\,\Omega \) and a capacitor of \(5.0\,\mu F.\) How much is the current \(I\) in the circuit in steady state? ................... \(A\)

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Answer & Solution

Correct Answer

(B) \(0.67\)

Step-by-step Solution

Detailed explanation

In steady state capacitor is fully charged hence no current will flow through line \(2\) .

Bysimplyfing the circuit

Hence resultant potential difference across resistances will be \(8.0 \,\mathrm{V}\)

Thus current \(1=\frac{V}{R}=\frac{8.0}{3+9}=\frac{8}{12}\)

or, \(\quad I=\frac{2}{3}=0.67\, \mathrm{A}\)

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