Question

For the circuit shown in the figure

• A. the current $I$ through the battery is $7.5mA$
• B. the potential difference across $R_L$ is $18V$
• 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)

$R_{total=2+\frac{6\times 1.5}{6+1.5}}=3.2k\Omega$
$(A)I=\frac{25V}{3.2k\Omega }=7.5mA=I_{R_1}$
$I_{R_2}=\left(\frac{R_L}{R_L+R_2}\right)I$
$I=\frac{1.5}{7.5}\times 7.5=1.5mA$
$I_{R_2}=6mA$
$(B)V_{R_L}=(I_{R_L})(R_L)=9V$
$(C)\frac{P_{R_1}}{P_{R_2}}=\frac{(I_{R_1}^2{R}_1)}{(I_{R_2}^2)R_2}=\frac{(7.5{)}^2(2)}{(1.5{)}^2(6)}=\frac{25}{3}$
(D) Now potential difference across $R_L$ will be $V_L=24\left[\frac{6/7}{6+6/7}\right]=3V$ Earlier it was 9 V
Since, $p=\frac{V^2}{R}orP=V^2$
In new situation potential difference has been decreased three times. Therefore, power dissipated will decrease by a factor of 9.