Question

Potential of point A is

• A. 20 V
• B. 10 V
• C. 9 V
• D. -9 V

Answer 1

Answer: (C)

9 V

Answer 2

To find the potential of point A, we first need to interpret the circuit diagram correctly.

1. Identify Ground: Point C is grounded, which means its potential $V_C = 0$ V.

2. Identify Connected Nodes:

   • The line connecting A and B indicates that points A and B are at the same potential. So, $V_A = V_B$.
   • The line connecting E and D indicates that points E and D are at the same potential. So, $V_E = V_D$.

3. Analyze the 1 Ω Resistor: The 1 Ω resistor is connected between points E and D. Since $V_E = V_D$, the potential difference across the 1 Ω resistor is $V_E - V_D = 0$. Therefore, no current flows through the 1 Ω resistor, and it is effectively shorted (i.e., it behaves like a wire, or it can be removed from the circuit without affecting the currents and voltages in other parts of the circuit).

4. Simplify the Circuit: Let's denote the common potential of A and B as $V_{AB}$ and the common potential of E and D as $V_{ED}$.

   • The 20 V battery is connected between A and E. The positive terminal is towards A. So, $V_A - V_E = 20$ V. This means $V_{AB} - V_{ED} = 20$ V.
   • The lower branch consists of the 9 Ω resistor and the 10 Ω resistor connected in series between B and D, with point C (grounded) in between. This branch is effectively connected between $V_{AB}$ (at B) and $V_{ED}$ (at D).
   • The total resistance of this lower branch is $R_{lower} = 9 \, \Omega + 10 \, \Omega = 19 \, \Omega$.

5. Calculate the Current: The potential difference across the lower branch (between B and D) is $V_B - V_D$. Since $V_B = V_{AB}$ and $V_D = V_{ED}$, we have $V_B - V_D = V_{AB} - V_{ED} = 20$ V. The current ($I$) flowing through the lower branch is given by Ohm's Law: $I = \frac{V_B - V_D}{R_{lower}} = \frac{20 \, \text{V}}{19 \, \Omega} = \frac{20}{19}$ A.

6. Calculate the Potential of Point A: We need to find $V_A$, which is the same as $V_B$. The current $I$ flows from B to C through the 9 Ω resistor. Using Ohm's law for the 9 Ω resistor: $V_B - V_C = I \times 9 \, \Omega$ We know $V_C = 0$ V and $I = \frac{20}{19}$ A. $V_B - 0 = \left(\frac{20}{19}\right) \times 9$ $V_B = \frac{180}{19}$ V

   Since $V_A = V_B$, we have: $V_A = \frac{180}{19}$ V

7. Approximate the Value: $\frac{180}{19} \approx 9.47$ V.

8. Compare with Options: The calculated value 9.47 V is closest to 9 V among the given options.