Physics Solution – Question 19

Solution to Question 19

1. Charging Phase Analysis

The circuit consists of a battery $V_0$ charging a capacitor $C$ through a resistor $R$. The device D behaves as an open switch (non-conducting) initially. The voltage across the capacitor $V_C(t)$ as a function of time during charging is given by the standard RC circuit equation:

$$ V_C(t) = V_0 \left( 1 – e^{-\frac{t}{RC}} \right) $$

2. Determining the Period

The device D remains non-conducting until the voltage across the capacitor rises to a threshold value $V_1$. Let $T$ be the time taken for the voltage to rise from 0 to $V_1$.

Setting $V_C(T) = V_1$:

$$ V_1 = V_0 \left( 1 – e^{-\frac{T}{RC}} \right) $$

Once $V_C$ reaches $V_1$, the device D rapidly discharges the capacitor to a negligibly small value. Since the discharge is “rapid,” the time taken for discharge is effectively zero compared to the charging time. Thus, the total time period of the oscillation is determined by the charging time $T$.

3. Mathematical Derivation

We solve the charging equation for $T$:

$$ \frac{V_1}{V_0} = 1 – e^{-\frac{T}{RC}} $$ $$ e^{-\frac{T}{RC}} = 1 – \frac{V_1}{V_0} $$ $$ e^{-\frac{T}{RC}} = \frac{V_0 – V_1}{V_0} $$

Taking the natural logarithm ($\ln$) on both sides:

$$ -\frac{T}{RC} = \ln \left( \frac{V_0 – V_1}{V_0} \right) $$ $$ T = -RC \ln \left( \frac{V_0 – V_1}{V_0} \right) $$

Using the property $-\ln(x) = \ln(1/x)$, we invert the argument of the logarithm:

$$ T = RC \ln \left( \frac{V_0}{V_0 – V_1} \right) $$

The voltage developed across the capacitor is periodic with time period: $$ RC \ln \left( \frac{V_0}{V_0 – V_1} \right) $$ (Option c)