JEE Challenger
More from Electrostatic Potential and Capacitance
PhyHardAdv 2016 • Paper 2

Transient Response in RC Bridge Circuit with Voltmeter and Ammeter

In the circuit diagram shown below, the switch is closed at time t=0t = 0. Which of the following statements is/are true?

Question Diagram 1

Options

A

The voltmeter displays −5 V-5\text{ V} as soon as the key is pressed, and displays +5 V+5\text{ V} after a long time

Correct
B

The voltmeter will display 0 V0\text{ V} at time t=ln⁡2 secondst = \ln 2\text{ seconds}

Correct
C

The current in the ammeter becomes 1/e1/e of the initial value after 1 second1\text{ second}

Correct
D

The current in the ammeter becomes zero after a long time

Correct

Topics & Concepts

Electrostatic Potential and CapacitanceCapacitorsKirchhoff's Laws

Step-by-Step Solution

To determine which of the statements are correct, let us analyze the transient behavior of the given RC bridge circuit after the key is closed at time t=0t = 0.

1. Circuit Parameters and Setup

Let:

  • The DC supply voltage be V=5 VV = 5\text{ V}.
  • The potential at the right vertex (connected to the negative terminal of the battery) be VR=0 VV_R = 0\text{ V}.
  • The potential at the left vertex (connected to the positive terminal of the battery through the ammeter) be VL=5 VV_L = 5\text{ V}.
  • The top node potential be Vtop(t)V_{\text{top}}(t).
  • The bottom node potential be Vbottom(t)V_{\text{bottom}}(t).

The voltmeter is connected with its positive terminal at the bottom node and negative terminal at the top node. Therefore, the reading displayed by the voltmeter is: Vm(t)=Vbottom(t)−Vtop(t)V_m(t) = V_{\text{bottom}}(t) - V_{\text{top}}(t)

2. Time Constants of the Branches

  • Top Branch: Consists of capacitor C1=40 μFC_1 = 40\ \mu\text{F} and resistor R1=25 kΩR_1 = 25\text{ k}\Omega. τ1=R1C1=(25×103 Ω)×(40×10−6 F)=1 s\tau_1 = R_1 C_1 = (25 \times 10^3\ \Omega) \times (40 \times 10^{-6}\text{ F}) = 1\text{ s}

  • Bottom Branch: Consists of resistor R2=50 kΩR_2 = 50\text{ k}\Omega and capacitor C2=20 μFC_2 = 20\ \mu\text{F}. τ2=R2C2=(50×103 Ω)×(20×10−6 F)=1 s\tau_2 = R_2 C_2 = (50 \times 10^3\ \Omega) \times (20 \times 10^{-6}\text{ F}) = 1\text{ s}

3. Node Potentials as Functions of Time tt

  • Top Node (Vtop(t)V_{\text{top}}(t)): At t=0+t = 0^+, the uncharged capacitor C1C_1 acts as a short circuit, so Vtop(0+)=VL=5 VV_{\text{top}}(0^+) = V_L = 5\text{ V}. At steady state (t→∞t \to \infty), C1C_1 acts as an open circuit, so no current flows through R1R_1, yielding Vtop(∞)=VR=0 VV_{\text{top}}(\infty) = V_R = 0\text{ V}. Thus, Vtop(t)=5e−t/τ1=5e−t VV_{\text{top}}(t) = 5 e^{-t/\tau_1} = 5 e^{-t}\text{ V}

  • Bottom Node (Vbottom(t)V_{\text{bottom}}(t)): At t=0+t = 0^+, the uncharged capacitor C2C_2 acts as a short circuit, so Vbottom(0+)=VR=0 VV_{\text{bottom}}(0^+) = V_R = 0\text{ V}. At steady state (t→∞t \to \infty), C2C_2 acts as an open circuit, so no current flows through R2R_2, yielding Vbottom(∞)=VL=5 VV_{\text{bottom}}(\infty) = V_L = 5\text{ V}. Thus, Vbottom(t)=5(1−e−t/τ2)=5(1−e−t) VV_{\text{bottom}}(t) = 5\left(1 - e^{-t/\tau_2}\right) = 5(1 - e^{-t})\text{ V}

4. Evaluation of Options

Option A & Option B: Voltmeter Reading

The voltmeter reading at time tt is: Vm(t)=Vbottom(t)−Vtop(t)=5(1−e−t)−5e−t=5(1−2e−t) VV_m(t) = V_{\text{bottom}}(t) - V_{\text{top}}(t) = 5(1 - e^{-t}) - 5e^{-t} = 5(1 - 2e^{-t})\text{ V}

  • Just after the key is pressed (t=0+t = 0^+): Vm(0+)=5(1−2)=−5 VV_m(0^+) = 5(1 - 2) = -5\text{ V}

  • After a long time (t→∞t \to \infty): Vm(∞)=5(1−0)=+5 VV_m(\infty) = 5(1 - 0) = +5\text{ V} Statement (A) is TRUE.

  • To find the time tt when Vm(t)=0 VV_m(t) = 0\text{ V}: 5(1−2e−t)=0  ⟹  2e−t=1  ⟹  et=2  ⟹  t=ln⁡2 seconds5(1 - 2e^{-t}) = 0 \implies 2e^{-t} = 1 \implies e^t = 2 \implies t = \ln 2\text{ seconds} Statement (B) is TRUE.

Option C & Option D: Ammeter Current

The total current supplied by the source and passing through the ammeter is I(t)=i1(t)+i2(t)I(t) = i_1(t) + i_2(t), where:

  • i1(t)i_1(t) is the current in the top branch: i1(t)=Vtop(t)−VRR1=5e−t25×103=0.2e−t mAi_1(t) = \frac{V_{\text{top}}(t) - V_R}{R_1} = \frac{5 e^{-t}}{25 \times 10^3} = 0.2 e^{-t}\text{ mA}
  • i2(t)i_2(t) is the current in the bottom branch: i2(t)=VL−Vbottom(t)R2=5−5(1−e−t)50×103=5e−t50×103=0.1e−t mAi_2(t) = \frac{V_L - V_{\text{bottom}}(t)}{R_2} = \frac{5 - 5(1 - e^{-t})}{50 \times 10^3} = \frac{5 e^{-t}}{50 \times 10^3} = 0.1 e^{-t}\text{ mA}

The total current measured by the ammeter is: I(t)=(0.2+0.1)e−t mA=0.3e−t mAI(t) = (0.2 + 0.1)e^{-t}\text{ mA} = 0.3 e^{-t}\text{ mA}

  • At t=0 st = 0\text{ s}, I(0)=0.3 mAI(0) = 0.3\text{ mA}.

  • At t=1 st = 1\text{ s}: I(1)=0.3e−1 mA=I(0)eI(1) = 0.3 e^{-1}\text{ mA} = \frac{I(0)}{e} Statement (C) is TRUE.

  • As t→∞t \to \infty: I(∞)=lim⁡t→∞0.3e−t=0 AI(\infty) = \lim_{t \to \infty} 0.3 e^{-t} = 0\text{ A} Statement (D) is TRUE.

Conclusion

Statements A, B, C, and D are all correct.

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