Analysis of Resistor Circuit with Load Resistance and Battery

To determine the correct statements, we analyze the circuit step-by-step:

1. Initial Circuit Analysis

From the circuit diagram:

• The battery voltage is $V = 24\text{ V}$.
• The resistors are $R_1 = 2\text{ k}\Omega$, $R_2 = 6\text{ k}\Omega$, and the load resistor is $R_{\text{L}} = 1.5\text{ k}\Omega$.
• The resistor $R_2$ is in parallel with the load resistor $R_{\text{L}}$, and this combination is connected in series with $R_1$.

The equivalent resistance of the parallel combination of $R_2$ and $R_{\text{L}}$ is:
$R_{2\parallel\text{L}} = \frac{R_2 R_{\text{L}}}{R_2 + R_{\text{L}}} = \frac{6 \times 1.5}{6 + 1.5} = \frac{9}{7.5} = 1.2\text{ k}\Omega$

The total equivalent resistance of the circuit connected to the battery is:
$R_{\text{eq}} = R_1 + R_{2\parallel\text{L}} = 2\text{ k}\Omega + 1.2\text{ k}\Omega = 3.2\text{ k}\Omega$

Current through the battery:

$I = \frac{V}{R_{\text{eq}}} = \frac{24\text{ V}}{3.2\text{ k}\Omega} = 7.5\text{ mA}$
Hence, Option (A) is correct.

Potential difference across $R_{\text{L}}$:

The voltage across $R_{\text{L}}$ is the same as the voltage across the parallel combination:
$V_{\text{L}} = I \times R_{2\parallel\text{L}} = 7.5\text{ mA} \times 1.2\text{ k}\Omega = 9\text{ V}$
Since $V_{\text{L}} = 9\text{ V} \neq 18\text{ V}$, Option (B) is incorrect.

Ratio of powers dissipated in $R_1$ and $R_2$:

The voltage across $R_1$ is:
$V_1 = V - V_{\text{L}} = 24\text{ V} - 9\text{ V} = 15\text{ V}$

The power dissipated in $R_1$ is:
$P_1 = \frac{V_1^2}{R_1} = \frac{15^2}{2\text{ k}\Omega} = \frac{225}{2}\text{ mW} = 112.5\text{ mW}$

The power dissipated in $R_2$ is:
$P_2 = \frac{V_{\text{L}}^2}{R_2} = \frac{9^2}{6\text{ k}\Omega} = \frac{81}{6}\text{ mW} = 13.5\text{ mW}$

The ratio of powers is:
$\frac{P_1}{P_2} = \frac{112.5}{13.5} = \frac{25}{3} \neq 3$
Hence, Option (C) is incorrect.

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2. When $R_1$ and $R_2$ are Interchanged

Now, the values are:

• $R_1' = 6\text{ k}\Omega$
• $R_2' = 2\text{ k}\Omega$
• $R_{\text{L}} = 1.5\text{ k}\Omega$

The new equivalent resistance of the parallel combination of $R_2'$ and $R_{\text{L}}$ is:
$R_{2'\parallel\text{L}} = \frac{R_2' R_{\text{L}}}{R_2' + R_{\text{L}}} = \frac{2 \times 1.5}{2 + 1.5} = \frac{3}{3.5} = \frac{6}{7}\text{ k}\Omega$

The total equivalent resistance becomes:
$R_{\text{eq}}' = R_1' + R_{2'\parallel\text{L}} = 6 + \frac{6}{7} = \frac{48}{7}\text{ k}\Omega$

Using the voltage divider formula, the new potential difference across $R_{\text{L}}$ is:
$V_{\text{L}}' = V \times \frac{R_{2'\parallel\text{L}}}{R_{\text{eq}}'} = 24\text{ V} \times \frac{\frac{6}{7}}{\frac{48}{7}} = 24 \times \frac{6}{48} = 3\text{ V}$

The ratio of the new power dissipated in $R_{\text{L}}$ to the initial power is:
$\frac{P_{\text{L}}'}{P_{\text{L}}} = \frac{\frac{(V_{\text{L}}')^2}{R_{\text{L}}}}{\frac{V_{\text{L}}^2}{R_{\text{L}}}} = \left(\frac{V_{\text{L}}'}{V_{\text{L}}}\right)^2 = \left(\frac{3\text{ V}}{9\text{ V}}\right)^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}$

Thus, the power dissipated in $R_{\text{L}}$ decreases by a factor of $9$.
Hence, Option (D) is correct.

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Conclusion

The correct options are A and D.