AP EAMCET · PHYSICS · Current Electricity
In steady state, a capacitor of capacitance \(2 \mu \mathrm{F}\) is charged to \(4 \mu \mathrm{C}\), as shown in figure. If the internal resistance of the cell is \(0.5 \Omega\), then the emf of the cell is
- A \(4 \mathrm{~V}\)
- B \(5 \mathrm{~V}\)
- C \(2.5 \mathrm{~V}\)
- D \(2 \mathrm{~V}\)
Answer & Solution
Correct Answer
(C) \(2.5 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
\[ \mathrm{V}_C=\frac{Q}{C}=\frac{4 \times 10^{-6}}{2 \times 10^{-6}}=2 \mathrm{v} \] Now, \(\mathrm{V}_C=\mathrm{V}\) (across \(2 \Omega\) resistor) \[ \therefore \quad I=\frac{V}{R}=\frac{2}{2}=1 \mathrm{~A} \] Now, using the relation, \(\mathrm{V}=E-I r\) So,…
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