KCET · Physics · Alternating Current
A capacitor of \( 8 \mathrm{~F} \) is connected as shown. Charge on the plates of the capacitor
- A \( 32 \) C
- B \( 40 \mathrm{C} \)
- C \( 00 \)
- D (\( 80 \mathrm{C} \)
Answer & Solution
Correct Answer
(A) \( 32 \) C
Step-by-step Solution
Detailed explanation
Current through \( 4 \Omega \) resister is \( I=\frac{E}{R+r}=\frac{5}{4+1}=1 A \)
Thus, potential difference across \( 4 \Omega \) resistor is
\(V=I R=1 A \times 4 \Omega=4 V\)
Therefore, charge on the plate of capacitor is
\(Q=C V=8 F \times 4 V=32 C\)
Thus, charge \( =32 \mathrm{C} \)
Thus, potential difference across \( 4 \Omega \) resistor is
\(V=I R=1 A \times 4 \Omega=4 V\)
Therefore, charge on the plate of capacitor is
\(Q=C V=8 F \times 4 V=32 C\)
Thus, charge \( =32 \mathrm{C} \)
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