Question

The final charge on the capacitor, when key \( S_1 \) is closed and \( S_2 \) is open, is: \includegraphics[width=0.5\linewidth]{PH30.1.png}
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• A. \(5 \, \mu C\)
• B. \(5 \, mC\)
• C. \(25 \, mC\)
• D. 0.1C
• A. Zero
• B. \(5 \, mC\)
• C. \(2.5 \, mC\)
• D. \(5 \, \mu C\)
• A. \( [M L^2 T^{-3} A^{-2}] \)
• B. \( [M^0 L^0 T^{-1} A^0] \)
• C. \( [M^{-1} L^{-2} T^4 A^2] \)
• D. \( [M^0 L^0 T^1 A^0] \)
• A. \( \frac{1}{2 \sqrt{e}} \, \text{mA} \)
• B. \( \sqrt{e} \, \text{mA} \)
• C. \( \frac{1}{\sqrt{e}} \, \text{mA} \)
• D. \( \frac{1}{2e} \, \text{mA} \)
• A. 5 \, mA
• B. 0.5 \, mA
• C. 2 \, mA
• D. 1 \, mA

Answer 1

The Correct Option is B

When key \( S_1 \) is closed and \( S_2 \) is open, the capacitor charges up fully and the final charge \( Q \) on the capacitor is given by: \[ Q = C \cdot V \] where \( C = 500 \, \mu F = 500 \times 10^{-6} \, F \) and \( V = 10 \, V \). Therefore: \[ Q = 500 \times 10^{-6} \times 10 = 5 \, mC \]

Answer 2

Once key \( S_1 \) is closed and \( S_2 \) is open, the capacitor is fully charged to the voltage \( V = 10 \, V \). When \( S_2 \) is closed, the charge on the capacitor remains the same, as it is disconnected from any other charge path. Hence, the final charge on the capacitor is \( 5 \, mC \).

Answer 3

The dimensional formula for \( RC \) can be derived as follows: - \( R \) (Resistance) has dimensions \( [M L^2 T^{-3} A^{-2}] \). - \( C \) (Capacitance) has dimensions \( [M^{-1} L^{-2} T^4 A^2] \). Thus, the dimensional formula for \( RC \) is: \[ [M L^2 T^{-3} A^{-2}] \]

Answer 4

The current in the resistor during the charging process of the capacitor is given by: \[ I(t) = \frac{V}{R} e^{-t/(RC)} \] After 5 seconds, the value of current will be: \[ I(5) = \frac{10}{20 \times 10^3} e^{-5/(RC)} \] where \( R = 20 \, k\Omega \) and \( C = 500 \, \mu F \). Solving gives the final result of current after 5 seconds as \( \frac{1}{\sqrt{e}} \, \text{mA} \).

Answer 5

The initial current when the capacitor begins charging is given by Ohm's Law: \[ I(0) = \frac{V}{R} \] where \( V = 10 \, V \) (battery voltage) and \( R = 20 \, k\Omega \). Therefore: \[ I(0) = \frac{10}{20 \times 10^3} = 0.5 \, \text{mA} \] Thus, the initial current is \( 0.5 \, \text{mA} \).