Question

In the given circuit, the potential at point $B$ is zero, the potential at points $A$ and $D$ will be:

• A. $V_A = 4 \ V; V_D = -9 \ V$
• B. $V_A = 3 \ V; V_D = -4 \ V$
• C. $V_A = 9 \ V; V_D = -3 \ V$
• D. $V_A = 4 \ V; V_D = -3 \ V$

Answer 1

Answer: (D)

(4)

Answer 2

The problem requires us to determine the potential at points A and D in the given circuit, given that the potential at point B is zero ($V_B = 0 \ V$). The current flowing through the circuit is 2 A from left to right.

1. Determine the potential at point A ($V_A$): The current flows from point A to point B through a 2 Ω resistor. According to Ohm's Law, the potential difference across the resistor is $V = I \times R$. Since current flows from higher potential to lower potential, $V_A$ must be greater than $V_B$. The potential drop across the 2 Ω resistor is $I \times R = 2 \ A \times 2 \ \Omega = 4 \ V$. Therefore, $V_A - V_B = 4 \ V$. Given $V_B = 0 \ V$, we have: $V_A - 0 = 4 \ V$ $V_A = 4 \ V$

2. Determine the potential at point C ($V_C$): The current flows from point B to point C through a 3 Ω resistor. The potential drop across the 3 Ω resistor is $I \times R = 2 \ A \times 3 \ \Omega = 6 \ V$. Since current flows from B to C, $V_B$ must be greater than $V_C$. So, $V_B - V_C = 6 \ V$. Given $V_B = 0 \ V$, we have: $0 - V_C = 6 \ V$ $V_C = -6 \ V$

3. Determine the potential at point D ($V_D$): There is a 3 V battery between points C and D. The negative terminal of the battery is connected to C, and the positive terminal is connected to D. This means that the potential at D is higher than the potential at C by 3 V. So, $V_D - V_C = 3 \ V$. Substitute the value of $V_C = -6 \ V$: $V_D - (-6 \ V) = 3 \ V$ $V_D + 6 \ V = 3 \ V$ $V_D = 3 \ V - 6 \ V$ $V_D = -3 \ V$

Thus, the potential at point A is $V_A = 4 \ V$ and the potential at point D is $V_D = -3 \ V$.