For the circuit shown in the figure:(A) The c | vedclass

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(B) Given: $V = 24 	ext{ V}$,$R_1 = 2 	ext{ k}\Omega$,$R_2 = 6 	ext{ k}\Omega$,$R_L = 1.5 	ext{ k}\Omega$.$1$. Calculate total resistance: $R_{	e

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$(A, D)$

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(B) Given: $V = 24 	ext{ V}$,$R_1 = 2 	ext{ k}\Omega$,$R_2 = 6 	ext{ k}\Omega$,$R_L = 1.5 	ext{ k}\Omega$.$1$. Calculate total resistance: $R_{	ext{total}} = R_1 + \frac{R_2 	imes R_L}{R_2 + R_L} = 2 + \frac{6 	imes 1.5}{6 + 1.5} = 2 + \frac{9}{7.5} = 2 + 1.2 = 3.2 	ext{ k}\Omega$.$2$. Current through battery: $I = \frac{V}{R_{	ext{total}}} = \frac{24 	ext{ V}}{3.2 	ext{ k}\Omega} = 7.5 	ext{ mA}$. (Statement $A$ is correct).$3$. Potential difference across $R_L$: $V_{R_L} = I 	imes R_{	ext{parallel}} = 7.5 	ext{ mA} 	imes 1.2 	ext{ k}\Omega = 9 	ext{ V}$. (Statement $B$ is incorrect).$4$. Power in $R_1$: $P_{R_1} = I^2 R_1 = (7.5)^2 	imes 2 = 56.25 	imes 2 = 112.5 	ext{ mW}$.Current in $R_2$: $I_{R_2} = I 	imes \frac{R_L}{R_2 + R_L} = 7.5 	imes \frac{1.5}{7.5} = 1.5 	ext{ mA}$.Power in $R_2$: $P_{R_2} = I_{R_2}^2 R_2 = (1.5)^2 	imes 6 = 2.25 	imes 6 = 13.5 	ext{ mW}$.Ratio: $\frac{P_{R_1}}{P_{R_2}} = \frac{112.5}{13.5} = 8.33 
eq 3$. (Statement $C$ is incorrect).$5$. Interchange $R_1$ and $R_2$: New $R_{	ext{total}} = 6 + \frac{2 	imes 1.5}{2 + 1.5} = 6 + \frac{3}{3.5} = 6 + 0.857 = 6.857 	ext{ k}\Omega$.New total current $I' = \frac{24}{6.857} \approx 3.5 	ext{ mA}$.New $V_{R_L} = I' 	imes R_{	ext{parallel}} = 3.5 	imes 0.857 = 3 	ext{ V}$.Power $P_L = \frac{V^2}{R_L}$. Since $V$ changes from $9 	ext{ V}$ to $3 	ext{ V}$ (factor of $3$),power changes by $3^2 = 9$. (Statement $D$ is correct).Thus,$(A)$ and $(D)$ are correct.

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