Question

The final charge on the capacitor, when key \( S_1 \) is closed and \( S_2 \) is open, is:

• 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. \( [M^0 L^0 T^1 A^0] \)
• B. \( [M^0 L^0 T^{-1} A^0] \)
• C. \( [M^{-1} L^{-2} T^4 A^2] \)
• D. \( [M L^2 T^{-3} A^{-2}] \)
• 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

For the RC series circuit with \( R = 20 \, k\Omega \), \( C = 500 \, \mu F \), and \( V = 10 \, V \), the charge \( q \) on the capacitor at time \( t \) when key \( S_1 \) is closed and \( S_2 \) is open is given by \( q = Q\left(1 - e^{-\frac{t}{RC}}\right) \). The maximum charge \( Q \) is \( Q = VC \). Calculating \( Q \):

\[ Q = VC = 10 \, V \cdot 500 \times 10^{-6} \, F = 5 \times 10^{-3} \, C \]

As \( t \to \infty \), the capacitor becomes fully charged. The final charge is therefore \( Q = 5 \, mC \).

The answer is \( 5 \, mC \).

Answer 2

To find the final charge on the capacitor when key S₁ is opened and S₂ is closed, we assume that the capacitor is fully charged. Initially, with S₁ closed and S₂ open, the capacitor charges until its potential difference equals the emf of the battery.

The charge on a fully charged capacitor is given by:

\[ Q = C V \]

Given:

• Resistance, \( R = 20\,\text{k}\Omega = 20{,}000\,\Omega \)
• Capacitance, \( C = 500\,\mu\text{F} = 500 \times 10^{-6}\,\text{F} \)
• Voltage, \( V = 10\,\text{V} \)

Substituting the values:

\[ Q = 500 \times 10^{-6} \times 10 = 5 \times 10^{-3}\,\text{C} \]

\[ Q = 5\,\text{mC} \]

Final Answer: The final charge on the capacitor when S₁ is opened and S₂ is closed is 5 mC.

Answer 3

The dimensional formula for \( RC \) is derived by multiplying the dimensional formulas for \( R \) (resistance) and \( C \) (capacitance).

1. Resistance \( R \):
\( R = \frac{V}{I} \). Given \( [V] = [M^1 L^2 T^{-3} A^{-1}] \) and \( [I] = [A^1] \), the dimensional formula for resistance is \( [R] = \frac{[M^1 L^2 T^{-3} A^{-1}]}{[A^1]} = [M^1 L^2 T^{-3} A^{-2}] \).

2. Capacitance \( C \):
\( C = \frac{Q}{V} \). Given \( [Q] = [A^1 T^1] \) and \( [V] = [M^1 L^2 T^{-3} A^{-1}] \), the dimensional formula for capacitance is \( [C] = \frac{[A^1 T^1]}{[M^1 L^2 T^{-3} A^{-1}]} = [M^{-1} L^{-2} T^4 A^2] \).

3. \( RC \) Product:
Multiplying the dimensions: \( [RC] = [R] \times [C] = [M^1 L^2 T^{-3} A^{-2}] \times [M^{-1} L^{-2} T^4 A^2] \).

Simplifying yields: \( [RC] = [M^{1-1} L^{2-2} T^{-3+4} A^{-2+2}] = [M^0 L^0 T^1 A^0] \).

Thus, the dimensional formula for \( RC \) is \( [M^0 L^0 T^1 A^0] \).

Answer 4

To determine the resistor current at 5 seconds, begin with the RC circuit charging current formula: \( i(t) = \frac{d}{dt}[q(t)] = \frac{d}{dt}\left[Q(1 - e^{-t/RC})\right] \). Differentiating \(q(t)\) with respect to \(t\) yields: \( i(t) = \frac{d}{dt}(Q - Qe^{-t/RC}) = Q \times \frac{d}{dt}(-e^{-t/RC}) \). This simplifies to: \[ i(t) = Q \cdot \left(-\frac{-1}{RC}e^{-t/RC}\right) \]. Therefore, the current expression is: \( i(t) = \frac{Q}{RC}e^{-t/RC} \). Given \( R = 20 \, \text{k}\Omega = 20000 \, \Omega \), \( C = 500 \, \mu\text{F} = 500 \times 10^{-6} \, \text{F} \), and \( V = 10 \, \text{V} \). The maximum capacitor charge \( Q \) is calculated as: \[ Q = C \times V = 500 \times 10^{-6} \times 10 \, \text{C} = 5 \times 10^{-3} \, \text{C} \]. The current expression \( i(t) \) is then: \( i(t) = \frac{5 \times 10^{-3}}{20000 \times 500 \times 10^{-6}}e^{-t/(20000 \times 500 \times 10^{-6})} \). Simplification results in: \( i(t) = \frac{5}{10000}e^{-t/(10)} \), which further simplifies to: \( i(t) = \frac{1}{2000}e^{-t/10} \). Substituting \( t = 5 \) seconds: \( i(5) = \frac{1}{2000}e^{-5/10} = \frac{1}{2000}e^{-0.5} \). Since \( \frac{1}{\sqrt{e}} = e^{-0.5} \): \( i(5) = \frac{1}{2000} \cdot \frac{1}{\sqrt{e}} = \frac{1}{\sqrt{e} \times 2000} \). Converting to mA: \( i(5) = \frac{1}{\sqrt{e} } \, \text{mA} \). Consequently, the resistor current at 5 seconds is \( \frac{1}{\sqrt{e}} \, \text{mA} \).

Answer 5

The initial charging current in the resistor of a given RC series circuit is determined by the formula for charging current \(I(t)\), derived from the charge equation: \(q = Q[1 - e^{-t/RC}]\). Current \(I(t)\) is the time derivative of charge \(q\), i.e., \(I(t) = \frac{dq}{dt}\). Differentiating the charge equation yields: \(I(t) = \frac{d}{dt}(Q[1 - e^{-t/RC}])\). Applying the derivative of an exponential function: \(I(t) = Q \cdot \frac{d}{dt} (1 - e^{-t/RC}) = Q \cdot \Big(0 + \frac{1}{RC} \cdot e^{-t/RC}\Big)\). Simplifying this expression results in: \(I(t) = \frac{Q}{RC} \cdot e^{-t/RC}\). At \(t = 0\), the exponential term \(e^{0}\) equals 1. Consequently, the initial current \(I(0)\) is: \(I(0) = \frac{Q}{RC}\). Substituting \(Q = VC\): \(I(0) = \frac{VC}{RC} = \frac{V}{R}\). With the given values \(R = 20 \, \text{k}\Omega = 20,000 \, \Omega\), \(C = 500 \, \mu\text{F} = 500 \times 10^{-6} \, \text{F}\), and \(V = 10 \, \text{V}\), the calculation is: \(I(0) = \frac{10}{20,000} = 0.0005 \, \text{A} = 0.5 \, \text{mA}\). The initial charging current in the resistor is thus 0.5 mA.