Question

In the figure, potential difference between A and B is:

• A. 10 V
• B. 5 V
• C. 15 V
• D. zero

Answer 1

Answer: (A)

10 V

Answer 2

Set the potential at point B to 0 V ($V_B = 0$). The voltage source is 30 V, so the potential at its positive terminal is 30 V. Let's denote the node before R1 as P, so $V_P = 30$ V. Applying KCL at node A: Current through R1 ($I_{R1}$) + Current through diode ($I_D$) = Current through R2 ($I_{R2}$) $\frac{V_P - V_A}{R1} + I_D = \frac{V_A - V_B}{R2}$ $\frac{30 - V_A}{10 \text{ kΩ}} + I_D = \frac{V_A - 0}{10 \text{ kΩ}}$ For an ideal diode, if it conducts, $V_A = V_D$ and $I_D > 0$. If it doesn't conduct, $I_D = 0$ and $V_A < V_D$. Assume the diode conducts: $V_A = V_D$. The current through R3 is $I_{R3} = \frac{V_D - V_B}{R3} = \frac{V_A - 0}{10 \text{ kΩ}}$. Since $I_D = I_{R3}$ (KCL at D), $I_D = \frac{V_A}{10 \text{ kΩ}}$. Substituting into the KCL at A: $\frac{30 - V_A}{10 \text{ kΩ}} + \frac{V_A}{10 \text{ kΩ}} = \frac{V_A}{10 \text{ kΩ}}$ $30 - V_A + V_A = V_A$ $30 = 2V_A \implies V_A = 15 \text{ V}$ If $V_A = 15$ V, then $V_D = 15$ V. $I_D = \frac{15 \text{ V}}{10 \text{ kΩ}} = 1.5$ mA. Check KCL at A: $I_{R1} = \frac{30-15}{10} = 1.5$ mA. $I_{R2} = \frac{15-0}{10} = 1.5$ mA. $1.5 \text{ mA} + 1.5 \text{ mA} \neq 1.5 \text{ mA}$. This assumption is incorrect.

Let's re-evaluate the circuit diagram. The voltage source is connected such that its positive terminal is at the top, and the negative terminal is at B. So, $V_{source} = V_{top} - V_{bottom} = 30$ V. If $V_B = 0$, then $V_{bottom} = 0$. This means $V_{top} = 30$ V. The node before R1 is $V_{top}$, so $V_P = 30$ V. KCL at node A: $\frac{V_P - V_A}{R1} = \frac{V_A - V_B}{R2} + I_{diode}$ $\frac{30 - V_A}{10 \text{ kΩ}} = \frac{V_A - 0}{10 \text{ kΩ}} + I_{diode}$ Assume diode conducts ($V_A = V_D$): $I_{diode} = \frac{V_D - V_B}{R3} = \frac{V_A - 0}{10 \text{ kΩ}}$. $\frac{30 - V_A}{10 \text{ kΩ}} = \frac{V_A}{10 \text{ kΩ}} + \frac{V_A}{10 \text{ kΩ}}$ $30 - V_A = V_A + V_A$ $30 = 3V_A \implies V_A = 10 \text{ V}$ If $V_A = 10$ V, then $V_D = 10$ V. $I_{diode} = \frac{10 \text{ V}}{10 \text{ kΩ}} = 1$ mA. This is $>0$, so the diode conducts. $I_{R1} = \frac{30-10}{10} = 2$ mA. $I_{R2} = \frac{10-0}{10} = 1$ mA. KCL at A: $I_{R1} = I_{R2} + I_{diode} \implies 2 \text{ mA} = 1 \text{ mA} + 1 \text{ mA}$. This is consistent. Therefore, $V_A = 10$ V. The potential difference between A and B is $V_{AB} = V_A - V_B = 10 \text{ V} - 0 \text{ V} = 10$ V.