Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Advanced 2014Paper 1
Exam
JEE Advanced
Year
2014
Paper
Paper 1
Subject
Physics
Chapter
Alternating Current

JEE Advanced 2014 · Paper 1 · Physics · Alternating Current

At time t = 0, terminal A in the circuit shown in the figure is...

At time t = 0, terminal A in the circuit shown in the figure is connected to B by a key and an alternating current II\left(t\right) = {I}_{0}cos (\omegat)\left(\omegat\right) , with I0=1A{I}_{0} = 1A and ω=500\omega = 500 rad/s starts flowing in it with the initial direction shown in the figure. At t=7π6ωt = \dfrac{7\pi}{6\omega} , the key is switched from B to D. Now onwards only A and D are connected. A total charge Q flows from the battery to charge the capacitor fully. If C=20\muF,R=10ΩC = 20\muF, R = 10Ω and the battery is ideal with of 50V, identify the correct statement(s).

At time t = 0, terminal A in the circuit shown in the figure is connected to B by a key and an alternating current I

Options

  1. AMagnitude of the maximum charge on the capacitor before t=7π6ωt = \dfrac{7\pi}{6\omega} is 1×10−3C1 \times {10}^{- 3}C .
  2. BThe current in the left part of the circuit just before t=7π6ωt = \dfrac{7\pi}{6\omega} is clockwise.
  3. CImmediately after A is connected to D, the current in R is 10 A.
  4. DQ=2×10−3CQ = 2 × {10}^{- 3}C .

Correct answer

C. Immediately after A is connected to D, the current in R is 10 A.

Step-by-step solution

Given, I = {I}_{0}\cos \left(\omega t\right), Peakormaximumcurrent{I}_{0} = 1Aangularfrequency, \omega = 500rad{s}^{- 1}capacitanceC = 20\mu FresistanceR = 10ΩbatteryvoltageV = 50VIf q represent the charge on capacitor&rsquo;s upper plate:<br />I\left(t\right) = {I}_{0}\cos ( \omegat )<br/><br />⇒ q\left(t\right) = \dfrac{{I}_{0}}{\omega}\sin \left(\omegat\right)<br/><br />∴Max charge \left({q}_\max\right) = \dfrac{1A}{500\dfrac{Rad}{s}} = 2 \times {10}^{- 3}C
Charge of upper plate at t=76ωt = \dfrac{7}{6\omega}
=I0ωsin⁡(7πω)=−I02ω= \dfrac{{I}_{0}}{\omega}\sin \left(\dfrac{7\pi}{\omega}\right) = - \dfrac{{I}_{0}}{2\omega}
When capacitor is fully charged then charge on upper plate =50v×20\muf= 50v \times 20\muf
=10×10−3C= 10 \times {10}^{- 3}C
⇒Q=1×10−3−(I02ω)=2×10−3C⇒ Q = 1 \times {10}^{- 3} - \left(\dfrac{{I}_{0}}{2\omega}\right) = 2 \times {10}^{- 3}C
Voltage across capacitor when A and D are connected =1×10−3C20×10−6\muf=50V= \dfrac{1 \times {10}^{- 3}C}{20 \times {10}^{- 6}\muf} = 50V
Total voltage across R=100VR = 100V
Current is anticlockwise
Practice Alternating Current on Quantrex Academy →

More from Alternating Current

Consider a circuit consisting of a capacitor of capacitance C and a coil with N turns per… JEE Advanced 2026A LCR series circuit driven with E_ rms = 90 V at frequency f_d = 30 Hz has resistance R… JEE Main 2026A series LCR circuit with R = 20\ Ω , L = 1.6 H and C = 40\ is connected to a variable… JEE Main 2026An a.c. source of angular frequency ω is connected across a resistor R and a capacitor C… JEE Main 2026An inductor of 10 mH, capacitor of 0.1\ μ F and a resistor of 100\ Ω are connected in… JEE Main 2026The figure given below shows an LCR series circuit with two switches S_1 and S_2 . When… JEE Main 2026An inductor stores 16 J of magnetic field energy and dissipates 32 W of thermal energy… JEE Main 2026The electric current in the circuit is given as i=i_ o (t / T) . The r.m.s current for… JEE Main 2026 Full Alternating Current list All JEE Advanced PYQs