A current of $$15 \text{ mA}$$ flows in the circuit as shown in figure. The value of potential difference between the points $$A$$ and $$B$$ will be

We need to find the potential difference between points $$A$$ and $$B$$ ($$V_A - V_B$$) for the given resistor network.

1. Calculate the Equivalent Resistance of the Network

Let's analyze the arrangement of the resistors from the circuit diagram 

• The top $$10\text{ k}\Omega$$ resistor and the middle $$5\text{ k}\Omega$$ resistor are connected in parallel with each other. Let's find their equivalent resistance ($$R_p$$):

  $$R_p = \frac{10 \times 5}{10 + 5} = \frac{50}{15} = \frac{10}{3}\text{ k}\Omega$$

• This parallel combination ($$R_p$$) is connected in series with the bottom-left $$5\text{ k}\Omega$$ resistor and the right vertical $$10\text{ k}\Omega$$ resistor.

Therefore, the total equivalent resistance ($$R_{\text{eq}}$$) of the entire circuit branch connected between terminals $$A$$ and $$B$$ is:

$$R_{\text{eq}} = 5\text{ k}\Omega + R_p + 10\text{ k}\Omega$$

$$R_{\text{eq}} = 15 + \frac{10}{3} = \frac{45 + 10}{3} = \frac{55}{3}\text{ k}\Omega$$

2. Calculate the Potential Difference ($$V_{AB}$$)

The total current flowing into the circuit at terminal $$A$$ and exiting at terminal $$B$$ is given as $$I = 15\text{ mA}$$.

Applying Ohm's law ($$V = I \times R$$) across the entire network between points $$A$$ and $$B$$:

$$V_{AB} = I \times R_{\text{eq}}$$

$$V_{AB} = 15\text{ mA} \times \frac{55}{3}\text{ k}\Omega$$

Since $$\text{mA} \times \text{k}\Omega = 10^{-3}\text{ A} \times 10^3\ \Omega = 1$$, the units simplify cleanly to volts:

$$V_{AB} = 15 \times \frac{55}{3} = 5 \times 55 = 275\text{ V}$$

Conclusion

The potential difference between points $$A$$ and $$B$$ is 275 V, which matches Option D.