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JEE Advanced 2010Paper 2
Exam
JEE Advanced
Year
2010
Paper
Paper 2
Subject
Physics
Chapter
Alternating Current

JEE Advanced 2010 · Paper 2 · Physics · Alternating Current

You are given many resistances, capacitors and inductors. These are...

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 Hz50 \mathrm{~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 ACA C ) 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 You are given many resistances, capacitors and inductors. These are connected to a variable DC voltage source (the

Options

  1. A(A) R,S,T, (B) Q,R,T, (C) P, (D) Q,R,S
  2. B(A) R,S,T, (B) Q,R,S,T, (C) P,Q, (D) Q,R,S,T
  3. C(A) S,T, (B) Q,R,T, (C) P,Q, (D) Q,R,S,T
  4. D(A) S,T, (B) Q,R,S,T, (C) P, (D) Q,R,S

Correct answer

B. (A) R,S,T, (B) Q,R,S,T, (C) P,Q, (D) Q,R,S,T

Step-by-step solution

In circuit (P), I can't be non zero in steady state. In circuit (Q) : V1=0andV2=2I=VV_1=0 and V_2=2 I=V (also)  In  circuit  (R): V1=XLI=(2\pif L)I=(2π×50×6×10−3)I=1.88IV2=2I\begin{aligned}& \text{\text{ In }\text{ circuit } (R): }V_1=X_L I=(2 \pif \mathrm{~L})I \\ & \quad=\left(2 \pi \times50 \times6 \times10^{-3}\right) I \\ & \quad=1.88 I \\ & V_2=2 I \end{aligned} In circuit (S):V_1=X_L I=1.88 I V2=XCI=(12πfC)I=(12π×50×3×10−6)I=(1061)I\begin{array}{r}V_2=X_C I=\left(\frac{1}{2 \pi f C}\right)I \\ =\left(\frac{1}{2 \pi \times 50 \times 3 \times 10^{-6}}\right)I=(1061) I \end{array} In circuit (T):V1=IR=(1000)IV2=XCI=(1061)I\begin{aligned}& V_1=I R=(1000) I \\ & V_2=X_C I=(1061) I \end{aligned} Thereforethecorrectoptionsareasunder(A)→R,S,Tthe correct options are as under (A) → R, S, T\quad (B) → Q, R, S, T (C) → Q or P,QP, Q \quad (D) → Q, R, S, T
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