Analysis of the Circuit

1. Identify the circuit topology: The circuit shown is a Wheatstone bridge. The battery with voltage $V = 30 \text{ V}$ is connected across the input terminals. The resistor $R_L$ is connected between the intermediate nodes of the two parallel branches.

2. List the component values:

   • $R_1 = 2 \text{ k}\Omega$
   • $R_2 = 4 \text{ k}\Omega$
   • $R_3 = 2 \text{ k}\Omega$
   • $R_4 = 4 \text{ k}\Omega$
   • $R_L = 8 \text{ k}\Omega$
   • $V = 30 \text{ V}$

3. Check for the balance condition: A Wheatstone bridge is balanced if the ratio of resistances in the opposite arms is equal. The condition is $\frac{R_1}{R_2} = \frac{R_3}{R_4}$.

   • Substituting the given values: $\frac{2 \text{ k}\Omega}{4 \text{ k}\Omega} = \frac{1}{2}$ and $\frac{2 \text{ k}\Omega}{4 \text{ k}\Omega} = \frac{1}{2}$.
   • Since $\frac{1}{2} = \frac{1}{2}$, the bridge is balanced.

4. Consequences of a balanced bridge: For a balanced bridge, the potential at the node between $R_1$ and $R_3$ (let's call it C) is equal to the potential at the node between $R_2$ and $R_4$ (let's call it D). Therefore, no current flows through the resistor $R_L$. The circuit can be simplified by removing $R_L$.

Evaluation of the Options

Option A: The current I through the battery is 7.5 mA.

1. With the bridge balanced, the circuit simplifies to two parallel branches.
   • Top branch resistance: $R_{top} = R_1 + R_3 = 2 \text{ k}\Omega + 2 \text{ k}\Omega = 4 \text{ k}\Omega$.
   • Bottom branch resistance: $R_{bottom} = R_2 + R_4 = 4 \text{ k}\Omega + 4 \text{ k}\Omega = 8 \text{ k}\Omega$.
2. The equivalent resistance of the circuit is the parallel combination of these two branches: $R_{eq} = \frac{R_{top} \times R_{bottom}}{R_{top} + R_{bottom}} = \frac{4 \times 8}{4 + 8} = \frac{32}{12} = \frac{8}{3} \text{ k}\Omega$
3. The total current $I$ from the battery is: $I = \frac{V}{R_{eq}} = \frac{30 \text{ V}}{(8/3) \text{ k}\Omega} = \frac{30 \times 3}{8 \times 10^3} \text{ A} = \frac{90}{8} \text{ mA} = 11.25 \text{ mA}$
4. The calculated current is 11.25 mA, not 7.5 mA. Therefore, Option A is incorrect.

Option B: The potential difference across $R_L$ is 18 V.

1. As established, the bridge is balanced. This means the potential difference across the central resistor $R_L$ is zero.
2. Therefore, Option B is incorrect.

Option C: Ratio of powers dissipated in $R_1$ and $R_2$ is 3.

1. The voltage across both parallel branches is 30 V.
2. Current through the top branch ($R_1, R_3$): $I_1 = \frac{V}{R_{top}} = \frac{30 \text{ V}}{4 \text{ k}\Omega} = 7.5 \text{ mA}$.
3. Current through the bottom branch ($R_2, R_4$): $I_2 = \frac{V}{R_{bottom}} = \frac{30 \text{ V}}{8 \text{ k}\Omega} = 3.75 \text{ mA}$.
4. Power dissipated in $R_1$: $P_1 = I_1^2 R_1 = (7.5 \times 10^{-3})^2 \times (2 \times 10^3) = 56.25 \times 10^{-6} \times 2 \times 10^3 = 112.5 \text{ mW}$.
5. Power dissipated in $R_2$: $P_2 = I_2^2 R_2 = (3.75 \times 10^{-3})^2 \times (4 \times 10^3) = 14.0625 \times 10^{-6} \times 4 \times 10^3 = 56.25 \text{ mW}$.
6. The ratio of powers is $\frac{P_1}{P_2} = \frac{112.5}{56.25} = 2$.
7. The option states the ratio is 3. Therefore, Option C is incorrect.

Option D: If $R_1$ and $R_2$ are interchanged, magnitude of the power dissipated in $R_L$ will decrease by a factor of 9.

1. In the original configuration, the bridge is balanced, so the power dissipated in $R_L$ is $P_L = 0$.
2. Now, interchange $R_1$ and $R_2$. The new values are $R'_1 = 4 \text{ k}\Omega$ and $R'_2 = 2 \text{ k}\Omega$. The other resistors remain the same: $R_3 = 2 \text{ k}\Omega$, $R_4 = 4 \text{ k}\Omega$.
3. Check the balance condition for the new circuit: $\frac{R'_1}{R'_2} = \frac{4}{2} = 2$. And $\frac{R_3}{R_4} = \frac{2}{4} = \frac{1}{2}$. Since $2 \neq \frac{1}{2}$, the new bridge is unbalanced.
4. Because the new bridge is unbalanced, a non-zero current will flow through $R_L$, and the power dissipated in it, $P'_L$, will be greater than zero.
5. The power changes from $P_L=0$ to $P'_L > 0$. This is an increase, not a decrease. The phrasing "decrease by a factor of 9" is nonsensical in this context.
6. Therefore, Option D is incorrect.

Conclusion

Based on a rigorous analysis of the circuit diagram and the values provided, none of the options are correct. There is likely an error in the problem statement (either the values of the components or the options provided) or the stored correct answer.

For instance, if we assume there is a typo in the resistance values such that $R_3=4\text{ k}\Omega$ and $R_4=8\text{ k}\Omega$, the bridge would still be balanced ($2/4 = 4/8$), and the equivalent resistance would be $R_{eq} = (R_1+R_3) || (R_2+R_4) = (2+4)||(4+8) = 6||12 = 4 \text{ k}\Omega$. This would make the total current $I = 30V / 4k\Omega = 7.5 \text{ mA}$, making option A correct. However, even with this change, option D's statement remains problematic as the power in $R_L$ would increase from zero upon interchanging $R_1$ and $R_2$.

Given the contradictions, and sticking strictly to the information provided, no option is valid.