Physics · Current Electricity

JEE Advanced 2022 — Paper 1 — Question 28

The figure shows a circuit having eight resistances of 1Ω1 \Omega each, labelled R1\mathrm{R}_{1} to R8\mathrm{R}_{8},

and two ideal batteries with voltages ε1=12 V\varepsilon_{1}=12 \mathrm{~V} and ε2=6 V\varepsilon_{2}=6 \mathrm{~V}.

Which of the following statement(s) is(are) correct?

Question figure
  1. Option A:

    The magnitude of current flowing through R1\mathrm{R}_{1} is 7.2 A .

    Correct
  2. Option B:

    The magnitude of current flowing through R2R_{2} is 1.2 A .

    Correct
  3. Option C:

    The magnitude of current flowing through R3R_{3} is 4.8 A .

    Correct
  4. Option D:

    The magnitude of current flowing through R5\mathrm{R}_{5} is 2.4 A .

    Correct

Answer: A, B, C, D

Step-by-step solution

−(i−i1)+6−12(i−i1)+12i1=0-\left(i-i_{1}\right)+6-\frac{1}{2}\left(i-i_{1}\right)+\frac{1}{2} i_{1}=0

−32(i−i1)+i12+6=0-\frac{3}{2}\left(i-i_{1}\right)+\frac{i_{1}}{2}+6=0

−32i+32i1+i22+6=0-\frac{3}{2} \mathrm{i}+\frac{3}{2} \mathrm{i}_{1}+\frac{\mathrm{i}_{2}}{2}+6=0

−32i+2i1+6=0-\frac{3}{2} \mathrm{i}+2 \mathrm{i}_{1}+6=0

−i12−i2−i+12=0-\frac{i_{1}}{2}-\frac{i}{2}-i+12=0 −i12−32i+12=0-\frac{\mathrm{i}_{1}}{2}-\frac{3}{2} \mathrm{i}+12=0

52i1−6=0\frac{5}{2} \mathrm{i}_{1}-6=0 i1=6×25=125=2.4\mathrm{i}_{1}=\frac{6 \times 2}{5}=\frac{12}{5}=2.4

−1.2−1.5i+12=0-1.2-1.5 \mathrm{i}+12=0 1.5i=10.81.5 \mathrm{i}=10.8 i=10.81.5i=\frac{10.8}{1.5}0

i=7.2\mathrm{i}=7.2

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2022
Paper
Paper 1
Subject
Physics
Chapter
Current Electricity
Topic
Circuit Analysis, Kirchhoff's Law and Nodal Analysis