Question

For the circuit shown in the figure

• A. the current $I$ through the battery is $7.5 \,mA$
• B. the potential difference across $R_L$ is $18\, V$
• C. ratio of powers dissipated in $R_1 $ and $R_2$ is $3$
• D. if $R_1 $ and $ R_2$ are interchanged, magnitude of the power dissipated in $R_L$ will decrease by a factor of $9$

Answer 1

Answer: (A), (D)

$R_{total = 2 + \frac{6 \, \times \, 1.5}{6 \, + \, 1.5} } = 3.2k \Omega$
$(A) \, I = \frac{25V}{3.2k \Omega} = 7.5mA = I_{R_1}$
$ \, \, \, \, \, \, \, \, I_{R_2} = \bigg( \frac{R_L}{ R_L + R_2} \bigg) I$
$ \, \, \, \, \, \, \, \, \, \, I = \frac{1.5}{7.5} \times 7.5 = 1.5mA$
$ \, \, \, \, \, \, I_{R_2} = 6mA$
$(B)V_{R_L} = (I_{R_L}) (R_L) = 9V$
$(C) \frac{P_{R_1}}{P_{R_2}} = \frac{(I^2_{R_1} R_1)}{(I^2_{R_2})R_2} = \frac{(7.5)^2 (2)}{(1.5)^2(6)} = \frac{25}{3}$
(D) Now potential difference across $R_L$ will be $V_L = 24 \bigg[ \frac{6/7}{6 + 6/7} \bigg] = 3V$ Earlier it was 9 V
Since, $\, \, \, \, \, \, \, p=\frac{V^2}{R} \, \, or \, \, P= V^2$
In new situation potential difference has been decreased three times. Therefore, power dissipated will decrease by a factor of 9.