EMI BYU 33 - CHATUP707

Solution to Problem 33

Solution: Circuit Analysis with Inductors and Capacitors

Part (a): Voltmeter Reading (Steady State)

First, we analyze the circuit a long time after the switch is closed (steady state). In this state:

Inductors behave as short circuits (zero resistance wires).
Capacitors behave as open circuits (infinite resistance).

Calculating Potentials:
Let the potential at the negative terminal of the battery be $0$ and the positive terminal be $\mathcal{E}$.

Center Node (O): Connected directly to the positive terminal via the bottom-left inductor (short circuit). Potential $V_O = \mathcal{E}$.
Top-Right Node (TR): Connected directly to the negative terminal via the right-vertical inductor (short circuit). Potential $V_{TR} = 0$.
Top-Left Node (TL): Connected to the Center node via resistor $3R$. Since the capacitor in the left branch is open, no current flows through $3R$. Therefore, $V_{TL} = V_O = \mathcal{E}$.

The voltmeter is connected between the Top-Left and Top-Right nodes.

$$ V_{(a)} = V_{TL} – V_{TR} = \mathcal{E} – 0 = \mathcal{E} $$

Part (b): Voltmeter Reading (Just After Opening Switch)

Now, consider the moment immediately after the switch is opened. The battery is disconnected, but the current in the inductors cannot change instantaneously.

Initial Currents (from steady state):

Left Inductor: Current flows from Bottom-Left to Center. $I = \frac{V_O – V_{TR}}{R} = \frac{\mathcal{E}}{R}$.
Right Inductor: Current flows from Top-Right to Bottom-Right. $I = \frac{\mathcal{E}}{R}$.

Analysis of Independent Loops:
When the switch opens, the circuit splits into two independent loops:

Left Loop: Current $\mathcal{E}/R$ flows: Center $\to$ $3R$ $\to$ TL $\to$ Capacitor $\to$ BL $\to$ Inductor $\to$ Center.
Right Loop: Current $\mathcal{E}/R$ flows: Center $\to$ $R$ $\to$ TR $\to$ Inductor $\to$ BR $\to$ Capacitor $\to$ Center.

Calculating Potentials relative to Center (O):

At Top-Left: Current flows from Center to TL through $3R$.
$V_{TL} = V_O – I(3R) = V_O – \left(\frac{\mathcal{E}}{R}\right)(3R) = V_O – 3\mathcal{E}$.
At Top-Right: Current flows from Center to TR through $R$.
$V_{TR} = V_O – I(R) = V_O – \left(\frac{\mathcal{E}}{R}\right)(R) = V_O – \mathcal{E}$.

The voltmeter reading is:

            $$ V_{(b)} = V_{TL} – V_{TR} = (V_O – 3\mathcal{E}) – (V_O – \mathcal{E}) = -2\mathcal{E} $$
        

Part (c): Heat Dissipated in Resistor R

After the switch is opened, all the energy stored in the components will eventually be dissipated as heat in the resistors.

Since the loops are independent, the heat dissipated in resistor $R$ comes solely from the energy stored in the Right Loop components at the moment of switching.

Energy in Right Loop Components:

Inductor (Right): Carrying current $I = \mathcal{E}/R$.
Energy $U_L = \frac{1}{2}L I^2 = \frac{1}{2}L \left(\frac{\mathcal{E}}{R}\right)^2 = \frac{L\mathcal{E}^2}{2R^2}$.
Capacitor (Diagonal): At steady state, potential difference across it was $V_O – V_{BR} = \mathcal{E} – 0 = \mathcal{E}$.
Energy $U_C = \frac{1}{2}C \mathcal{E}^2$.

Total heat dissipated in $R$:

$$ H_R = \frac{C\mathcal{E}^2}{2} + \frac{L\mathcal{E}^2}{2R^2} $$