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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$(A, D)$

Vedclass · Answer

(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.

Similar Questions

The total power dissipated in the circuit shown here is ............. $W$.

The cold junction of a thermocouple is at $0^{\circ} C$. The thermo e.m.f. produced in the thermocouple is given by the equation $E = 16T - 0.04T^2$, where $T$ is the temperature of the hot junction. The temperature of inversion and the neutral temperature of the thermocouple are:

If the internal resistance of a cell is proportional to the current drawn from the cell,then the best representation of the terminal potential difference of the cell with respect to the current drawn from the cell will be:

$A$ student measures the terminal potential difference $(V)$ of a cell (of $emf \; \varepsilon$ and internal resistance $r$) as a function of the current $(I)$ flowing through it. The slope and intercept of the graph between $V$ and $I$,respectively,are equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module