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← IIT-JEE 2010 · Paper 2
PhysicsMatrix match+8Question 57 of 57

IIT-JEE 2010 · Paper 2, Physics question 57

You are given many resistances, capacitors and inductors. These are connected to a variable DC voltage source (the first two circuits) or an AC voltage source of 50 Hz frequency (the next three circuits) in different ways as shown in Column II. When a current II (steady state for DC or rms for AC) flows through the circuit, the corresponding voltage V1V_1 and V2V_2 (indicated in circuits) are related as shown in Column I. Match the two.

Column I

  • A
    I≠0I\ne0, V1V_1 is proportional to II
  • B
    I≠0I\ne0, V2>V1V_2>V_1
  • C
    V1=0V_1=0, V2=VV_2=V
  • D
    I≠0I\ne0, V2V_2 is proportional to II

Column II

  • P
    Variable DC source V across a 6 mH inductor (voltage V₁) in series with a 3 μF capacitor (voltage V₂).
  • Q
    Variable DC source V across a 6 mH inductor (voltage V₁) in series with a 2 Ω resistor (voltage V₂).
  • R
    AC source V (50 Hz) across a 6 mH inductor (voltage V₁) in series with a 2 Ω resistor (voltage V₂).
  • S
    AC source V (50 Hz) across a 6 mH inductor (voltage V₁) in series with a 3 μF capacitor (voltage V₂).
  • T
    AC source V (50 Hz) across a 1 kΩ resistor (voltage V₁) in series with a 3 μF capacitor (voltage V₂).
Mark every match for each entry
PQRST
A
B
C
D
Answer and worked solution

Official answer: A → R, S, T; B → Q, R, S, T; C → P, Q; D → Q, R, S, T

Steady DC. An inductor has no voltage across it; a capacitor blocks current.

  • P (L + C on DC): I=0I=0, V1=0V_1=0, V2=VV_2=V.
  • Q (L + 2 Ω on DC): I=V/2I=V/2, V1=0V_1=0, V2=V=2IV_2=V=2I.

AC at 50 Hz. XL=2π×50×6×10−3≈1.9 ΩX_L=2\pi\times50\times6\times10^{-3}\approx1.9\ \Omega, XC=12π×50×3×10−6≈1061 ΩX_C=\dfrac1{2\pi\times50\times3\times10^{-6}}\approx1061\ \Omega.

  • R (L + 2 Ω): V1=1.9IV_1=1.9I, V2=2IV_2=2I.
  • S (L + C): V1=1.9IV_1=1.9I, V2=1061IV_2=1061I.
  • T (1 kΩ + C): V1=1000IV_1=1000I, V2=1061IV_2=1061I.

Matching.

  • A (I≠0I\ne0, V1∝IV_1\propto I): R, S, T. (In Q, V1=0V_1=0.)
  • B (I≠0I\ne0, V2>V1V_2>V_1): Q, R, S, T.
  • C (V1=0V_1=0, V2=VV_2=V): P, Q.
  • D (I≠0I\ne0, V2∝IV_2\propto I): Q, R, S, T.

Answer: A → R, S, T; B → Q, R, S, T; C → P, Q; D → Q, R, S, T