Step 1: Calculate the Average Voltage $\langle V_2 \rangle$

The graph shows the variation of voltage $V_2$ across the resistor $R_2$ with respect to time.

• From $t = 0$ to $t = 2\, \mu\text{s}$, $V_2 = 30\, \text{V}$.
• From $t = 2$ to $t = 3\, \mu\text{s}$, $V_2 = 0\, \text{V}$.
• The cycle repeats after $t = 3\, \mu\text{s}$, so the time period $T = 3\, \mu\text{s}$.

The average value of $V_2$ over one complete cycle is given by the area under the $V$-$t$ graph divided by the time period:

$$ \langle V_2 \rangle = \frac{\int_0^T V_2(t) dt}{T} $$ $$ \langle V_2 \rangle = \frac{(30\, \text{V} \times 2\, \mu\text{s}) + (0\, \text{V} \times 1\, \mu\text{s})}{3\, \mu\text{s}} = \frac{60}{3} = 20\, \text{V} $$

Step 2: Apply Current Conservation for Average Values

Let $i_{total}(t)$ be the current flowing from the source. This current flows through the parallel combination of $R_1$ and $C$, and then through $R_2$.

At any instant, the current through $R_2$ is $i_2(t) = i_{R1}(t) + i_C(t)$.

Taking the average over one time period:

$$ \langle i_2 \rangle = \langle i_{R1} \rangle + \langle i_C \rangle $$

Key Physics Principle: In a DC steady state (or periodic steady state where fluctuations are negligible), the average current flowing through a capacitor is zero ($\langle i_C \rangle = 0$).

Therefore:

$$ \langle i_2 \rangle = \langle i_{R1} \rangle $$

Step 3: Calculate the Average Voltage Across $R_1$

Using Ohm’s Law for the average currents:

$$ \langle i_2 \rangle = \frac{\langle V_2 \rangle}{R_2} \quad \text{and} \quad \langle i_{R1} \rangle = \frac{\langle V_{R1} \rangle}{R_1} $$

Equating the two expressions:

$$ \frac{\langle V_2 \rangle}{R_2} = \frac{\langle V_{R1} \rangle}{R_1} $$

We need to find the average voltage across $R_1$, denoted as $\langle V_{R1} \rangle$. Rearranging the equation:

$$ \langle V_{R1} \rangle = \langle V_2 \rangle \times \frac{R_1}{R_2} $$

Substitute the known values ($R_1 = 10\, \text{k}\Omega$, $R_2 = 5\, \text{k}\Omega$, and $\langle V_2 \rangle = 20\, \text{V}$):

$$ \langle V_{R1} \rangle = 20\, \text{V} \times \frac{10\, \text{k}\Omega}{5\, \text{k}\Omega} $$ $$ \langle V_{R1} \rangle = 20 \times 2 = 40\, \text{V} $$