JEE Mains · Physics · STD 12 - 3. current electricity
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? ................... \(A\)
- A \(1.6\)
- B \(0.67\)
- C \(2.5\)
- D \(0.25\)
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,…
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