JEE Advanced · Physics · 19. Current Electricity

For the circuit shown in the figure

A the current $I$ through the battery is $7.5 \mathrm{~mA}$
B the potential difference across $R_L$ 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 $R_L$ will decrease by a factor of 9.

Verified Solution

Answer & Solution

Correct Answer

(D) if $R_1$ and $R_2$ are interchanged, magnitude of the power dissipated in $R_L$ will decrease by a factor of 9.

Step-by-step Solution

Detailed explanation

$R_{\text {total }}=2+\frac{6 \times 1.5}{6+1.5}=3.2 \mathrm{k} \Omega$
(A) $I=\frac{24 \mathrm{~V}}{3.2 \mathrm{k} \Omega}=7.5 \mathrm{~mA}=I_{R_1}$ $I_{R_2}=$ $\left(\frac{R_L}{R_L+R_2}\right) I$ $I=\frac{1.5}{7.5} \times 7.5=$ $1.5 \mathrm{~mA}$
(B) $V_{R_L}=\left(I_{R_L}\right)\left(R_L\right)=9 \mathrm{~V}$
(C) $\frac{P_{R_1}}{P_{R_2}}=\frac{\left(I_{R_1}^2\right) R_1}{\left(I_{R_2}^2\right) R_2}=\frac{(7.5)^2(2)}{(1.5)^2(6)}=\frac{25}{3}$
(D) When $R_1$ and $R_2$ are interchanged, then
$\frac{R_2 R_L}{R_2+R_L}=\frac{2 \times 1.5}{3.5}=\frac{6}{7} \mathrm{k} \Omega$
Now potential difference across $R_L$ will be
$V_L=24\left[\frac{6 / 7}{6+6 / 7}\right]$
Earlier it was $9 \mathrm{~V}$
Since, $\quad P=\frac{V^2}{R}$ 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.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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JEE Advanced 2013 Easy

Match List I with List II and select the correct answer using the codes given below the lists:

	$\text{List I}$		$\text{List II}$
a.	Boltzmann constant	p.	$\left[M L^2 T^{-1}\right]$
b.	Coefficient of viscosity	q.	$\left[M L^{-1} T^{-1}\right]$
c.	Planck constant	r.	$\left[M L T^{-3} K^{-1}\right]$
d.	Thermal conductivity	s.	$\left[M L^2 T^{-2} K^{-1}\right]$

JEE Advanced 2013 Easy

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	\(\text{List I}\)		\(\text{List II}\)
a.	Boltzmann constant	p.	\(\left[M L^2 T^{-1}\right]\)
b.	Coefficient of viscosity	q.	\(\left[M L^{-1} T^{-1}\right]\)
c.	Planck constant	r.	\(\left[M L T^{-3} K^{-1}\right]\)
d.	Thermal conductivity	s.	\(\left[M L^2 T^{-2} K^{-1}\right]\)