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

To find the final charge on the capacitor when key \( S_1 \) is closed and \( S_2 \) is open, we consider the provided RC series circuit. With the given values: \( R = 20 \, k\Omega \), \( C = 500 \, \mu F \), and \( V = 10 \, V \), the formula for the charge \( q \) on the capacitor over time \( t \) is:

\[ q = Q\left(1 - e^{-\frac{t}{RC}}\right) \]

Where \( Q \) is the maximum charge, given by \( Q = VC \). To find \( Q \):

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

Thus, the final charge on the capacitor, when fully charged (as \( t \to \infty \)), is:

\( Q = 5 \, mC \)

Therefore, the correct answer is \( 5 \, mC \).

Answer 2

To determine the final charge on the capacitor when key \( S_1 \) is opened and \( S_2 \) is closed, we consider the scenario after the capacitor has been fully charged. Initially, with \( S_1 \) closed and \( S_2 \) open, charges flow between the battery and the capacitor, charging the capacitor over time until the voltage across it equals the battery's emf. The charge on the capacitor \( Q \) is given by \( Q = C \cdot V \). In this circuit: 

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

Using the formula for the full charge on a capacitor, \( Q = C \cdot V \), we calculate: \[ Q = 500 \times 10^{-6} \, F \times 10 \, V = 5 \times 10^{-3} \, C = 5 \, mC \] Therefore, the final charge on the capacitor, when \( S_1 \) is opened and \( S_2 \) is closed, is 5 mC.

Answer 3

To determine the dimensional formula for \( RC \), we need to find the dimensions of both \( R \) (resistance) and \( C \) (capacitance), then multiply them.

1. Dimensional Formula for Resistance \( R \):
Resistance is given by the formula \( R = \frac{V}{I} \), where \( V \) (voltage) has a dimensional formula of \([M^1 L^2 T^{-3} A^{-1}]\) and \( I \) (current) has a dimensional formula of \([A^1]\).

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

2. Dimensional Formula for Capacitance \( C \):
Capacitance is given by the formula \( C = \frac{Q}{V} \), where \( Q \) (charge) has a dimensional formula of \([A^1 T^1]\) and \( V \) has a dimensional formula of \([M^1 L^2 T^{-3} A^{-1}]\).

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

3. Dimensional Formula for \( RC \):
Now, multiply the dimensional formulas of resistance and capacitance:
\([RC] = [R] \times [C] = [M^1 L^2 T^{-3} A^{-2}] \times [M^{-1} L^{-2} T^4 A^2]\)

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

Therefore, the dimensional formula for \( RC \) is \( [M^0 L^0 T^1 A^0] \). This is the correct answer from the provided options.

Answer 4

To find the value of current in the resistor after 5 seconds, we start by using the expression for the charging current in an RC circuit: \[ i(t) = \frac{d}{dt}[q(t)] = \frac{d}{dt}\left[Q(1 - e^{-t/RC})\right] \] Differentiate \(q(t)\) with respect to time \(t\): \( i(t) = \frac{d}{dt}(Q - Qe^{-t/RC}) = Q \times \frac{d}{dt}(-e^{-t/RC}) \) \[ i(t) = Q \cdot \left(-\frac{-1}{RC}e^{-t/RC}\right) \] Thus, the expression for current becomes: \( i(t) = \frac{Q}{RC}e^{-t/RC} \) Given values are \( R = 20 \, \text{k}\Omega = 20000 \, \Omega \), \( C = 500 \, \mu\text{F} = 500 \times 10^{-6} \, \text{F} \), and \( V = 10 \, \text{V} \). The charges on the capacitor, \( Q \), at full charge is: \[ Q = C \times V = 500 \times 10^{-6} \times 10 \, \text{C} = 5 \times 10^{-3} \, \text{C} \] The expression for current \( i(t) \) becomes: \( i(t) = \frac{5 \times 10^{-3}}{20000 \times 500 \times 10^{-6}}e^{-t/(20000 \times 500 \times 10^{-6})} \) Simplifying: \( i(t) = \frac{5}{10000}e^{-t/(10)} \) \( i(t) = \frac{1}{2000}e^{-t/10} \) Substituting \( t = 5 \) seconds: \( i(5) = \frac{1}{2000}e^{-5/10} \) \( i(5) = \frac{1}{2000}e^{-0.5} \) Given that \( \frac{1}{\sqrt{e}} = e^{-0.5} \): \( i(5) = \frac{1}{2000} \cdot \frac{1}{\sqrt{e}} \) \( i(5) = \frac{1}{\sqrt{e} \times 2000} \) Converting to mA: \( i(5) = \frac{1}{\sqrt{e} } \, \text{mA} \) So, the value of current in the resistor after 5 seconds is: \( \frac{1}{\sqrt{e}} \, \text{mA} \)

Answer 5

To calculate the initial charging current in the resistor for the given RC series circuit, we need to use the formula for charging current \(I(t)\), which is derived from the equation:

\(q = Q[1 - e^{-t/RC}]\)

The current \(I(t)\) is the rate of change of charge \(q\) with respect to time \(t\), hence \(I(t) = \frac{dq}{dt}\). Differentiating the charge equation, we have:

\(I(t) = \frac{d}{dt}(Q[1 - e^{-t/RC}])\)

Using the derivative of an exponential function, we can write:

\(I(t) = Q \cdot \frac{d}{dt} (1 - e^{-t/RC}) = Q \cdot \Big(0 + \frac{1}{RC} \cdot e^{-t/RC}\Big)\)

Simplifying, this gives us:

\(I(t) = \frac{Q}{RC} \cdot e^{-t/RC}\)

Initially, at \(t = 0\), the exponential term becomes \(e^{0} = 1\). Therefore, the initial current \(I(0)\) is:

\(I(0) = \frac{Q}{RC}\)

Since \(Q = VC\), substituting this into the expression gives:

\(I(0) = \frac{VC}{RC} = \frac{V}{R}\)

Given \(R = 20 \, \text{k}\Omega = 20,000 \, \Omega\), \(C = 500 \, \mu\text{F} = 500 \times 10^{-6} \, \text{F}\), and \(V = 10 \, \text{V}\), we substitute these values:

\(I(0) = \frac{10}{20,000} = 0.0005 \, \text{A} = 0.5 \, \text{mA}\)

Therefore, the initial value of the charging current in the resistor is 0.5 mA.