Question

In the circuit shown, all batteries are ideal and the values of the resistances are:

$R_1 = 10 \Omega$

$R_2 = 10 \Omega$

$R_3 = 5 \Omega$

Which of the following option(s) is/are correct?

• A. Current through $R_1$ is 2A from $D$ to $E$
• B. Current through $R_2$ is 1A from $F$ to $C$
• C. Potential difference ($V_F - V_C$) is 30 volts
• D. Maximum power dissipation will be in resistor $R_3$

Answer 1

None of the options are correct. The correct values are:

• Current through $R_1$ is 5A from E to D.
• Current through $R_2$ is 4A from F to C.
• Potential difference ($V_F - V_C$) is 40 volts.
• Maximum power dissipation will be in resistor $R_1$.

Answer 2

Here's how to analyze the circuit and determine the correct answers:

1. Node Potentials:

   • Assume the potential at node A is $V_A = 0V$ (ground).
   • Since the 30V battery is connected between H and A, with H being the positive terminal, $V_H = 30V$.
   • The points H, G, F, and E are connected by a wire, so they are at the same potential: $V_G = V_F = V_E = 30V$.
   • The 10V battery is connected between G and B, with G being the positive terminal, so $V_B = V_G - 10V = 30V - 10V = 20V$.
   • The 40V battery is connected between F and C, with F being the positive terminal, so $V_C = V_F - 40V = 30V - 40V = -10V$.
   • The 50V battery is connected between E and D, with E being the positive terminal, so $V_D = V_E - 50V = 30V - 50V = -20V$.

2. Currents through Resistors:

   • $R_1$: $I_{R_1} = (V_E - V_D) / R_1 = (30V - (-20V)) / 10\Omega = 5A$ (from E to D).
   • $R_2$: $I_{R_2} = (V_F - V_C) / R_2 = (30V - (-10V)) / 10\Omega = 4A$ (from F to C).
   • $R_3$: $I_{R_3} = (V_G - V_B) / R_3 = (30V - 20V) / 5\Omega = 2A$.

3. Power Dissipation:

   • $P_{R_1} = I_{R_1}^2 * R_1 = (5A)^2 * 10\Omega = 250W$.
   • $P_{R_2} = I_{R_2}^2 * R_2 = (4A)^2 * 10\Omega = 160W$.
   • $P_{R_3} = I_{R_3}^2 * R_3 = (2A)^2 * 5\Omega = 20W$.

Therefore, $R_1$ dissipates the maximum power.