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JEE Advanced 2009Paper 1

Exam: JEE Advanced
Year: 2009
Paper: Paper 1
Subject: Physics
Chapter: Current Electricity

JEE Advanced 2009 · Paper 1 · Physics · Current Electricity

For the circuit shown in the figure

For the circuit shown in the figure

For the circuit shown in the figure

Options

Athe current $I$ through the battery is $7.5 \mathrm{~mA}$
Bthe potential difference across $R_L$ is $18 \mathrm{~V}$
Cratio of powers dissipated in $R_1$ and $R_2$ is 3
Dif $R_1$ and $R_2$ are interchanged, magnitude of the power dissipated in $R_L$ will decrease by a factor of 9.

Correct answer

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

Step-by-step solution

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) II=\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.

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