Question

A uncharged capacitor of capacitance $C$ is connected with cell of emf $V$ through three resistors as shown. At $t =$ 0, the switch is closed and the potential difference across the plate of capacitor is measured 40% of cell emf at time $\frac{3RC \ln(\beta)}{2}$ sec. Find $\beta$

Answer 1

5

Answer 2

To solve this problem, we need to determine the Thevenin equivalent circuit seen by the capacitor. This involves finding the steady-state voltage across the capacitor ($V_{Th}$) and the equivalent resistance seen by the capacitor ($R_{Th}$).

1. Determine the Thevenin Voltage ($V_{Th}$):

In the steady state (when the capacitor is fully charged), the capacitor acts as an open circuit, meaning no current flows through the branch containing the capacitor.

Let's label the nodes. Let the bottom wire be ground (0V). The negative terminal of the cell is at 0V, so the positive terminal is at $V$.

When the switch is closed, the point after the switch is at potential $V$.

The circuit then consists of the top-left resistor $R$ and the middle vertical resistor $R$ in series, connected across the cell $V$. The top-right resistor $R$ is effectively out of the current path because the capacitor branch is open.

The current flowing through the series combination is:

$I_{ss} = \frac{V}{R + R} = \frac{V}{2R}$

The potential at the node between the top-left $R$ and the middle $R$ (which is also the node before the top-right $R$) is the potential drop across the middle $R$ (relative to ground):

$V_{node} = I_{ss} \times R = \frac{V}{2R} \times R = \frac{V}{2}$

Since no current flows through the top-right $R$, the potential across the capacitor terminals is the same as this node potential.

Thus, the Thevenin voltage (final voltage across the capacitor) is:

$V_{Th} = \frac{V}{2}$

2. Determine the Thevenin Resistance ($R_{Th}$):

To find $R_{Th}$, we deactivate the independent voltage source (replace the cell with a short circuit) and look into the terminals where the capacitor is connected.

When the voltage source $V$ is shorted, the point where the positive terminal of $V$ was connected is now connected to ground (0V).

Looking from the capacitor terminals (between the top-right $R$ and ground):

The top-right resistor $R$ is in series with the equivalent resistance of the rest of the circuit.

The other two resistors (top-left $R$ and middle $R$) are now connected in parallel between the common node and ground.

The equivalent resistance of these two parallel resistors is:

$R_{parallel} = \frac{R \times R}{R + R} = \frac{R^2}{2R} = \frac{R}{2}$

This $R_{parallel}$ is in series with the top-right resistor $R$.

Therefore, the Thevenin resistance is:

$R_{Th} = R + R_{parallel} = R + \frac{R}{2} = \frac{3R}{2}$

3. Apply the Capacitor Charging Equation:

The voltage across a charging capacitor in an RC circuit is given by:

$V_C(t) = V_{Th} (1 - e^{-t/R_{Th}C})$

Substitute $V_{Th} = \frac{V}{2}$ and $R_{Th} = \frac{3R}{2}$:

$V_C(t) = \frac{V}{2} \left(1 - e^{-t / (\frac{3R}{2}C)}\right)$

$V_C(t) = \frac{V}{2} \left(1 - e^{-2t / (3RC)}\right)$

4. Solve for $\beta$ using the given condition:

We are given that at time $t = \frac{3RC \ln(\beta)}{2}$ sec, the potential difference across the capacitor is 40% of the cell emf $V$.

So, $V_C(t) = 0.40 V = \frac{2}{5} V$.

Substitute these values into the charging equation:

$\frac{2}{5} V = \frac{V}{2} \left(1 - e^{-2t / (3RC)}\right)$

Divide both sides by $V$:

$\frac{2}{5} = \frac{1}{2} \left(1 - e^{-2t / (3RC)}\right)$

Multiply by 2:

$\frac{4}{5} = 1 - e^{-2t / (3RC)}$

Rearrange the equation to isolate the exponential term:

$e^{-2t / (3RC)} = 1 - \frac{4}{5} = \frac{1}{5}$

Now, substitute the given time $t = \frac{3RC \ln(\beta)}{2}$:

$e^{-2 \left( \frac{3RC \ln(\beta)}{2} \right) / (3RC)} = \frac{1}{5}$

Simplify the exponent:

$e^{- (3RC \ln(\beta)) / (3RC)} = \frac{1}{5}$

$e^{-\ln(\beta)} = \frac{1}{5}$

Using the property $e^{-\ln(x)} = \frac{1}{x}$:

$\frac{1}{\beta} = \frac{1}{5}$

Therefore, $\beta = 5$.

The final answer is $\boxed{5}$.

Explanation of the solution:

1. Find $V_{Th}$: In steady state, the capacitor acts as an open circuit. The circuit reduces to a voltage divider with the two resistors in series (top-left R and middle R). The voltage across the middle resistor is $V/2$, which is the steady-state voltage across the capacitor. So, $V_{Th} = V/2$.
2. Find $R_{Th}$: Short the voltage source. The capacitor sees the top-right resistor R in series with the parallel combination of the other two resistors. The parallel combination is $R/2$. So, $R_{Th} = R + R/2 = 3R/2$.
3. Use charging equation: The capacitor voltage is $V_C(t) = V_{Th}(1 - e^{-t/R_{Th}C})$. Substitute $V_{Th} = V/2$ and $R_{Th} = 3R/2$.
4. Solve for $\beta$: Given $V_C(t) = 0.4V = 2V/5$ at $t = \frac{3RC \ln(\beta)}{2}$. Substitute these into the charging equation: $\frac{2V}{5} = \frac{V}{2}(1 - e^{-2t/(3RC)})$ $\frac{4}{5} = 1 - e^{-2t/(3RC)}$ $e^{-2t/(3RC)} = \frac{1}{5}$ Substitute $t$: $e^{-2(\frac{3RC \ln(\beta)}{2})/(3RC)} = \frac{1}{5}$ $e^{-\ln(\beta)} = \frac{1}{5}$ $\frac{1}{\beta} = \frac{1}{5} \implies \beta = 5$.