The circuit shown in fig consists of the following \(E_{1}=6, E_{2}=2, E_{3}=3\) Volt \(\mathrm{R}_{1}=6, \mathrm{R}_{4}=3 \mathrm{Ohm}\) \(\mathrm{R}_{3}=4, \mathrm{R}_{2}=2 \mathrm{Ohm}\) $\mathrm{C}=5 \mu \mathrm{F}$$E_{1}=6 \mathrm{~V} \quad E_{2}=2 \mathrm{~V} \quad E_{3}=3 \mathrm{~V} \quad \mathrm{R}_{1}=6 \Omega$ \(R_{2}=2 \Omega \quad R_{3}=4 \Omega \quad R_{4}=3 \Omega\) The energy stored in the capacitor is. (A) \(4.8 \times 10^{-6} \mathrm{~J}\) (B) \(9.6 \times 10^{-6} \mathrm{~J}\) (C) \(1.44 \times 10^{-5} \mathrm{~J}\) (D) \(1.92 \times 10^{-5} \mathrm{~J}\)

We have three voltage sources (E1, E2, and E3), four resistors (R1, R2, R3, and R4), and one capacitor (C). The energy stored in the capacitor depends on the total voltage across the capacitor, so our main goal is to find that voltage.

To begin analyzing the circuit, we need to find the currents flowing through the resistors. Let's assume I1 is the current through R1, I2 is the current through R2, I3 is the current through R3, I4 is the current through R4, and Ic is the current through the capacitor. By applying Kirchhoff's Current Law, we can obtain the following equations: For node a: I1 + I2 = I3 For node b: I3 - Ic = I4

To find the voltages across resistors, we need to form a loop equation by applying Kirchhoff's Voltage Law. We will start at the positive terminal of E1 and move clockwise through the loop containing E1, E2, R1, and R3, back to the starting point. The loop equation becomes: E1 - E2 - I1 * R1 - I3 * R3 = 0

We now have 2 equations from Kirchhoff's Current Law and 1 equation from Kirchhoff's Voltage Law. Let's solve these equations for the currents: We can solve the loop equation for I1: I1 = (E1 - E2 - I3 * R3) / R1 Now substitute this into the node a equation and solve for I3: I3 = I2 + (E1 - E2 - I2 * R1 - I3 * R3) / R1 Rearranging the equation, we get: I3 * (R1 + R3) = (E1 - E2) * R1 - I2 * (R1 * R3) Now, substitute the values given in the problem: I3 * 10 = 24 - 12 * I2 Now, we can find the value of I2 by considering the nodes and the law of conservation of charge. We know that Ic = 0, as the capacitor is fully charged and at steady-state. Therefore: I2 = I4 Using Ohm's Law, we can then find I2: I2 = (E3 - I3 * R3) / R4 Now, substitution of the values: I2 = (3 - 4 * I3) / 3 We now have two equations for I2: I2 = (24 - 12 * I3) / 10 I2 = (3 - 4 * I3) / 3 Equating the two expressions, we get: (24 - 12 * I3) / 10 = (3 - 4 * I3) / 3

Solving the above equation, we get: I3 = 0.90 A Now we can find I2: I2 = (3 - 4 * 0.90) / 3 = 0.3 A I4 = I2 = 0.3 A And thus, I1 = (6 - 2 - 0.9 * 4) / 6 = 0.6 A

The voltage across the capacitor will be the same as the voltage across R2, as they are connected in parallel. Using Ohm's Law: V_R2 = I2 * R2 = 0.3 A * 2 Ω = 0.6 V V_Capacitor = V_R2 = 0.6 V

The energy stored in a capacitor can be calculated using the following formula: Energy = (1/2) * C * V_Capacitor^2 Substituting the given values: Energy = 0.5 * 5 x 10^(-6) F * (0.6 V)^2 = 9.0 x 10^(-7) J The energy stored in the capacitor is 9.0 x 10^(-7) J, which is closest to the value given in option (A). Therefore, the correct answer is (A) \(4.8 \times 10^{-6} \mathrm{~J}\).