Animated Solution for Physics - Current Electricity: Calculate the steady state current in the $2\Omega$ resistor shown in the circuit (see figure). The internal resistance of the battery is negligible and the capacitance of the condenser $C$ is $0.2\mu \text{F}$.

Let's dive into this interesting circuit problem. We are presented with a $6\text{ V}$ battery connected to a network that includes several resistors and a capacitor. Our primary objective is to determine the steady-state current flowing through the $2\Omega$ resistor located at the top of the parallel branches.

The most crucial piece of information in the problem statement is the term steady state. In a direct current (DC) circuit, a capacitor initially allows current to flow as it charges up. However, once it reaches its steady state, it becomes fully charged and acts exactly like an open circuit.

This means that the branch containing the capacitor $C$ and the $4\Omega$ resistor will have absolutely zero current flowing through it ($I_C=0$). Because of this, we can effectively remove this entire bottom branch from our analysis, simplifying the circuit significantly.

With the capacitor branch out of the picture, our circuit is reduced to a simple series-parallel combination. We have the $2\Omega$ and $3\Omega$ resistors connected in parallel. This parallel combination is then connected in series with the $2.8\Omega$ resistor and the $6\text{ V}$ battery.

First, let's find the equivalent resistance of the parallel section ($R_p$). Using the product-over-sum rule:

Now that we have the equivalent resistance of the parallel part, we can find the total resistance of the entire circuit by adding the series resistor:

Using Ohm's Law ($V=IR$), we can calculate the total current $i$ drawn from the battery:

Current Division

This total current of $1.5\text{A}$ flows through the $2.8\Omega$ resistor and then splits into the two parallel branches. To find the specific current flowing through the $2\Omega$ resistor ($i_2$), we apply the current divider rule.

The current through one branch of a two-branch parallel circuit is equal to the total current multiplied by the resistance of the other branch, divided by the sum of the resistances:

Substituting our values:

And there we have it! The steady-state current through the $2\Omega$ resistor is exactly $0.9\text{A}$.