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Physics · Capacitors and R-C Circuits

JEE Advanced 2019 — Paper 1 — Question 9

In the circuit shown, initially there is no charge on capacitors and keys S1S_{1} and S2S_{2} are open. The values of the

capacitors are C1=10μ F,C2=30μ F\mathrm{C}_{1}=10 \mu \mathrm{~F}, \mathrm{C}_{2}=30 \mu \mathrm{~F} and C3\mathrm{C}_{3} =C4=80μ F=\mathrm{C}_{4}=80 \mu \mathrm{~F}. Which of the statement(s) is/are

correct?

Question figure
  1. Option A:

    The key S1S_{1} is kept closed for long time such that capacitors are fully charged. Now key S2S_{2} is closed, at this time, the instantaneous current across 30Ω30 \Omega resistor (between points P and Q ) will be 0.2 A (round off to 1st 1^{\text {st }} decimal place).

  2. Option B:

    If key S1\mathrm{S}_{1} is kept closed for long time such that capacitors are fully charged, the voltage across the capacitor C1\mathrm{C}_{1} will be 4 V .

    Correct
  3. Option C:

    At time t=0t=0, the key S1S_{1} is closed, the instantaneous current in the closed circuit will be 25 mA

    Correct
  4. Option D:

    If key S1S_{1} is kept closed for long time such that capacitors are fully charged, the voltage difference between points P and Q will be 10 V .

Answer: B, C

Step-by-step solution

S1S_{1} closed for long time

figure

V10μ F=(4040+10)5=4 V\mathrm{V}_{10 \mu \mathrm{~F}}=\left(\frac{40}{40+10}\right) 5=4 \mathrm{~V}

VP−VQ=4VV_{P}-V_{Q}=4 V

⇒t=0\Rightarrow \mathrm{t}=0, key S is closed

i=570+100+30=25 mA\mathrm{i}=\frac{5}{70+100+30}=25 \mathrm{~mA}

figure

−70i1+30i2−6=0-70 i_{1}+30 i_{2}-6=0

−30i1−60i2+6=0-30 i_{1}-60 i_{2}+6=0

⇒i2=0.11 A\Rightarrow \mathrm{i}_{2}=0.11 \mathrm{~A}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2019
Paper
Paper 1
Subject
Physics
Chapter
Capacitors and R-C Circuits
Topic
Charging and Discharging of R-C Circuits

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