For the circuit shown in the figure

$(A)$ the current $I$ through the battery is $7.5 \mathrm{~mA}$

$(B)$ the potential difference across $R_{\mathrm{L}}$, is $18 \mathrm{~V}$

$(C)$ 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 $R_{\mathrm{L}}$ will decrease by a factor of $9$

Question

For the circuit shown in the figure

$(A)$ the current $I$ through the battery is $7.5 \mathrm{~mA}$

$(B)$ the potential difference across $R_{\mathrm{L}}$, is $18 \mathrm{~V}$

$(C)$ 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 $R_{\mathrm{L}}$ will decrease by a factor of $9$

IIT 2009, Diffcult

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Solution

$R_{\text {total }}=2+\frac{6 \times 1.5}{6+1.5}=3.2 k \Omega$

$(a)$ $I =\frac{24 V}{3.2 k \Omega}=7.5 mA= I _{R_1}$

$I _{R_2}=\left(\frac{R_L}{R_L+R_2}\right) I$

$I =\frac{1.5}{7.5} \times 7.5=1.5 mA$

$I _{ R _{ L }}=6 mA$

$(b)$ $V_{R_L}=\left(I_{R_1}\right)\left(R_L\right)=9 V$

$(c)$ $\frac{ P _{R_1}}{ P _{ R _2}}=\frac{ I _{ R _1}^2 R _1}{ I _{ R _2}^2 R _2}=\frac{(7.5)^2(2)}{(1.5)^2(6)}=\frac{25}{3}$

$(d)$ When $R _1$ and $R _2$ are inter changed then

$\frac{ R _2 R _{ L }}{ R _2+ R _{ L }}=\frac{2 \times 1.5}{3.5}=\frac{6}{7} k \Omega$

Now potential difference across $R _{ L }$ will be

$V _{ L }=24\left[\frac{6 / 7}{6+6 / 7}\right]=3 V$

Earlier it was $9 V$

Since, $P =\frac{ V ^2}{ R }$ or $P \propto V ^2$

In new situation potential difference has been decreased three times. Therefore, power dissipated will decrease by a factor of $9$ .

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