Animated Solution for Physics - Electromagnetic Induction: 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(t)=I0cos(ωt), with I0=1 A and ω=500 rad s−1 starts flowing in it with the initial direction shown in the figure. At t=6ω7π, 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=20μF, R=10Ω and the battery is ideal with emf of 50 V, identify the correct statement(s).
Select Answer:
* Multiple Correct
Visualized Solution
I(t)=I0cos(ωt)
Initial state: Switch A connected to B.
Circuit is an AC source with a capacitor C.
I(t)=I0cos(ωt)
I0=1 A, ω=500 rad/s
Qmax=ωI0
I=dtdQ⟹Q(t)=∫I0cos(ωt)dt=ωI0sin(ωt)
Qmax=ωI0=5001 C=2×10−3 C
Option (a) claims it is 1×10−3 C, so (a) is incorrect.
I(6ω7π)<0
At t=6ω7π, the phase is ωt=67π.
I(6ω7π)=I0cos(67π)=1×(−23)<0
Negative current implies anti-clockwise direction.
Option (b) claims it is clockwise, so (b) is incorrect.
Q(6ω7π)=−1 mC
Q(6ω7π)=5001sin(67π)
Q=2×10−3×(−21)=−1×10−3 C=−1 mC
The top plate has −1 mC and the bottom plate has +1 mC.
VC=−50 V
Switch connects A to D.
Right loop is active with C, R, and 50V battery.
Initial charge on top plate Q1=−1 mC.
Voltage across capacitor VC=CQ1=20μF−1 mC=−50 V (top relative to bottom).
50+50−IR=0
Applying KVL clockwise:
+50 V (battery) +50 V (capacitor) −IR=0
100−I(10)=0⟹I=10 A
Option (c) is correct.
Qf=+1 mC
In steady state, current I=0.
Final charge on top plate Qf=C×Vbattery=20μF×50V=1000μC=+1 mC.
Qflow=2 mC
Total charge flown Qflow=Qf−Qi
Qflow=1 mC−(−1 mC)=2 mC
Qflow=2×10−3 C
Option (d) is correct.
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The Sigma Insight: Alternating Current (AC) and Voltage
Solution Diagram
Analyzing the Setup
Imagine a circuit that lives a double life. Initially, it behaves as a pure AC circuit, but with the flick of a switch, it transforms into a DC transient circuit.
This is exactly what we are dealing with here. At t=0, the switch A is connected to B. This creates a closed loop on the left side, consisting of an AC source and a capacitor C.
The current flowing through this loop is given by I(t)=I0cos(ωt), where I0=1 A and ω=500 rad/s.
The AC Phase
Charge and Current
To understand the state of the capacitor, we need to find the charge on it. Current is the rate of flow of charge, mathematically expressed as I=dtdQ.
By integrating the current function, we can find the charge as a function of time:
Q(t)=∫I0cos(ωt)dt=ωI0sin(ωt)
The maximum charge Qmax occurs when the sine function reaches its peak value of 1.
Qmax=ωI0=5001=2×10−3 C
This immediately tells us that option (a) is incorrect, as it claims the maximum charge is 1×10−3 C.
Next, let's determine the direction of the current just before the switch is thrown at t=6ω7π.
Substituting this time into our current equation gives:
I(6ω7π)=I0cos(ω⋅6ω7π)=cos(67π)
Since 67π lies in the third quadrant, its cosine is negative (−23).
The problem defines the initial clockwise direction as positive. Therefore, a negative current implies that the current is flowing in the anti-clockwise direction. This makes option (b) incorrect.
The Moment of Switching
To analyze the DC phase, we must know the exact initial conditions. What is the charge on the capacitor at the exact moment the switch is thrown?
Let's evaluate our charge function at t=6ω7π:
Q=5001sin(67π)=2×10−3×(−21)=−1×10−3 C=−1 mC
This means the top plate of the capacitor has a charge of −1 mC, and the bottom plate has +1 mC.
The DC Transient Phase
At t=6ω7π, the switch connects A to D. The left loop is broken, and the right loop becomes active.
This new loop contains the capacitor, a 10Ω resistor, and a 50 V battery.
Let's apply Kirchhoff's Voltage Law (KVL) to find the initial current I immediately after switching. We will trace the loop in the clockwise direction.
Starting from the bottom right and going up through the battery, we gain 50 V.
Next, we go down through the capacitor. We are moving from the negative top plate (−1 mC) to the positive bottom plate (+1 mC).
The potential difference across the capacitor is VC=CQ=20μF1 mC=50 V. Since we are moving from negative to positive, this is a potential gain of +50 V.
Finally, moving through the resistor gives a potential drop of −IR.
Our KVL equation becomes:
+50 V+50 V−IR=0
100=10I⟹I=10 A
The current immediately after switching is indeed 10 A, making option (c) correct!
Steady State and Total Charge
As time passes, the capacitor will charge up and eventually block all DC current. In this steady state, the current becomes zero.
The capacitor is now fully charged by the 50 V battery. The final charge on the top plate (which is connected to the positive terminal of the battery) will be:
Qf=C×V=20μF×50 V=1000μC=+1 mC
The problem asks for the total charge Q that flows from the battery to fully charge the capacitor.
This is simply the difference between the final charge and the initial charge on the top plate:
Qflow=Qf−Qi=1 mC−(−1 mC)=2 mC=2×10−3 C
Thus, option (d) is also correct. This problem is a masterful test of tracking initial conditions across changing circuit topologies!
