Animated Solution for Physics - Current Electricity: In the circuit shown below, the key is pressed at time $t=0$. Which of the following statement(s) is (are) true?

Imagine you are an electron standing at the positive terminal of a $5\text{ V}$ battery. You have two paths to choose from. The top path leads you through a $40\mu \text{F}$ capacitor and a $25\text{k}\Omega$ resistor. The bottom path takes you through a $50\text{k}\Omega$ resistor and a $20\mu \text{F}$ capacitor. Bridging these two paths is an ideal voltmeter, silently observing the potential difference between node B (on the bottom path) and node A (on the top path).

This setup is a classic parallel RC circuit disguised as a Wheatstone bridge. To conquer it, we must understand how capacitors behave at the exact moment the switch is closed, and how they evolve over time.

The moment the key is pressed ($t=0$), the uncharged capacitors are completely empty. They offer absolutely zero resistance to the sudden rush of charge. In circuit terms, an uncharged capacitor acts exactly like a short circuit (a plain wire).

Because the capacitors are shorted, the voltage drop across them is zero. - For the top branch, node A is directly connected to the positive terminal of the battery. Thus, $V_A=5\text{ V}$. - For the bottom branch, node B is directly connected to the negative terminal of the battery. Thus, $V_B=0\text{ V}$.

The voltmeter measures the potential of B relative to A:

This immediately confirms that Option (a) is correct for the initial state!

Now, let's fast forward to a long time later. The capacitors have soaked up all the charge they can hold. They are now fully charged and refuse to let any more direct current pass through. They act as open circuits.

With no current flowing ($i_1=0$ and $i_2=0$), there is no voltage drop across the resistors (since $V=IR=0$). - For the top branch, node A is now at the same potential as the negative terminal. Thus, $V_A=0\text{ V}$. - For the bottom branch, node B is at the same potential as the positive terminal. Thus, $V_B=5\text{ V}$.

Let's check the voltmeter again:

This perfectly completes the verification of Option (a).

To understand the in-between moments, we need the time constants ($\tau =RC$) for both branches. - Top branch: ${\tau }_1=R_1{C}_1=(25\times 10^3\Omega )\times (40\times 10^{-6}\text{F})=1\text{ s}$ - Bottom branch: ${\tau }_2=R_2{C}_2=(50\times 10^3\Omega )\times (20\times 10^{-6}\text{F})=1\text{ s}$

Beautifully, both branches decay at the exact same rate! The currents as a function of time are:

Using these currents, we can track the potentials at nodes A and B:

When will the voltmeter read zero? We simply set $V_B(t)-V_A(t)=0$:

Taking the natural logarithm of both sides gives $t=\ln 2\text{seconds}$. This proves Option (b) is correct!

The ammeter sits on the main line, so it measures the total current supplied by the battery, which is the sum of the branch currents:

The initial current is $i(0)=0.3\text{mA}$. After exactly $1\text{ second}$, the current is $i(1)=0.3e^{-1}=\frac{i(0)}{e}$. This confirms Option (c).

Finally, as $t\to \infty$, the exponential term $e^{-t}$ approaches zero, meaning the total current $i(\infty )$ becomes zero. This confirms Option (d).

Every single statement is a beautiful reflection of the physics of RC circuits!