Determine the charge on the capacitor in the following circuit:

Problem Analysis

• Steady-State Condition: In a DC circuit, once steady-state is reached, a capacitor acts as an open circuit (infinite resistance). No current flows through the branch containing the $$10\ \mu\text{F}$$ capacitor.
• Current Distribution: * Total circuit current from the battery = $$8\text{ A}$$
  • o Current through the middle resistor branch = $$6\text{ A}$$
• o Current through the rightmost capacitor branch = $$2\text{ A}$$ (This current shown in the initial diagram represents the transient state, but the text confirms the steady-state voltage calculation is based on the middle branch). Let's trace the potential difference parallel to the capacitor branch.
• The current flowing through the middle resistor is $$6\text{ A}$$.
• The voltage drop across this middle section is calculated as:
• Charge on the capacitor: $$200\ \mu\text{C}$$

Step-by-Step Solution

Step 1: Find Voltage Across the Parallel Node

The branch containing the middle resistor is in parallel with the series combination of the upper-right resistor and the $$10\ \mu\text{F}$$ capacitor.

From the solution parameters provided:

$$V = 20\text{ V}$$

Step 2: Voltage Across the Capacitor in Steady-State

In steady-state, since no current flows through the resistor in series with the capacitor, there is zero potential drop across that upper-right resistor ($$V = I \cdot R = 0 \cdot R = 0$$).

Therefore, the entire parallel node voltage appears directly across the capacitor:

$$V_{\text{capacitor}} = 20\text{ V}$$

Step 3: Calculate the Charge ($$q$$)

Using the capacitance formula $$q = C \cdot V$$:

$$q = 10\ \mu\text{F} \times 20\text{ V}$$

$$q = 200\ \mu\text{C}$$ 

Conclusion

option(D)