JEE Challenger
More from Electromagnetic Induction
PhyHardAdv 2016 • Paper 1

Ratio of Maximum to Minimum Current in Parallel Inductor Circuit

Two inductors L1L_1 (inductance 1 mH1\ \text{mH}, internal resistance 3 Ω3\ \Omega) and L2L_2 (inductance 2 mH2\ \text{mH}, internal resistance 4 Ω4\ \Omega), alongside a resistor RR (resistance 12 Ω12\ \Omega) are connected in parallel across a 5 V5\ \text{V} battery. The circuit is turned on at instant t=0t = 0. Determine the ratio of the maximum to the minimum current (Imax/IminI_{\text{max}} / I_{\text{min}}) drawn from the battery source.

Official Numerical Answer8

Topics & Concepts

Electromagnetic InductionSelf and Mutual InductanceAC Circuits

Step-by-Step Solution

To find the ratio of the maximum to the minimum current (Imax/IminI_{\text{max}} / I_{\text{min}}) drawn from the battery, we analyze the behavior of each parallel branch in the circuit at t=0+t = 0^+ (just after the switch is closed) and at t→∞t \to \infty (steady state).

The three branches connected in parallel across the DC voltage source V=5 VV = 5\ \text{V} are:

  1. Pure resistor branch with resistance R=12 ΩR = 12\ \Omega.
  2. First inductor branch with inductance L1=1 mHL_1 = 1\ \text{mH} and internal resistance r1=3 Ωr_1 = 3\ \Omega.
  3. Second inductor branch with inductance L2=2 mHL_2 = 2\ \text{mH} and internal resistance r2=4 Ωr_2 = 4\ \Omega.

1. Current through each branch as a function of time:

  • Resistor branch (RR): Since there is no inductance in this branch, the current instantaneously reaches its steady-state value: IR(t)=VR=512 Afor all t>0I_R(t) = \frac{V}{R} = \frac{5}{12}\ \text{A} \quad \text{for all } t > 0

  • First inductor branch (L1,r1L_1, r_1): Due to self-inductance, the current grows exponentially: I1(t)=Vr1(1−e−t/τ1),where τ1=L1r1I_1(t) = \frac{V}{r_1} \left(1 - e^{-t/\tau_1}\right), \quad \text{where } \tau_1 = \frac{L_1}{r_1}

  • Second inductor branch (L2,r2L_2, r_2): Similarly, the current grows exponentially: I2(t)=Vr2(1−e−t/τ2),where τ2=L2r2I_2(t) = \frac{V}{r_2} \left(1 - e^{-t/\tau_2}\right), \quad \text{where } \tau_2 = \frac{L_2}{r_2}

The total current drawn from the battery at any time t>0t > 0 is: I(t)=IR(t)+I1(t)+I2(t)I(t) = I_R(t) + I_1(t) + I_2(t)

Since I1(t)I_1(t) and I2(t)I_2(t) are strictly increasing functions of time, I(t)I(t) increases monotonically from t=0+t = 0^+ to t→∞t \to \infty.

2. Minimum Current (IminI_{\text{min}}):

At t=0+t = 0^+, an inductor opposes any sudden change in current, acting as an open circuit (I1(0+)=0I_1(0^+) = 0 and I2(0+)=0I_2(0^+) = 0). Thus, the minimum current is drawn at t=0+t = 0^+: Imin=I(0+)=IR(0+)+I1(0+)+I2(0+)=VR+0+0I_{\text{min}} = I(0^+) = I_R(0^+) + I_1(0^+) + I_2(0^+) = \frac{V}{R} + 0 + 0 Imin=512 AI_{\text{min}} = \frac{5}{12}\ \text{A}

3. Maximum Current (ImaxI_{\text{max}}):

At t→∞t \to \infty, the inductors offer no opposition to steady direct current and act as zero-resistance wires (short circuits). The current in each branch is limited only by its resistance: Imax=I(∞)=IR(∞)+I1(∞)+I2(∞)=VR+Vr1+Vr2I_{\text{max}} = I(\infty) = I_R(\infty) + I_1(\infty) + I_2(\infty) = \frac{V}{R} + \frac{V}{r_1} + \frac{V}{r_2}

Substituting the given values: Imax=V(1R+1r1+1r2)=5(112+13+14)I_{\text{max}} = V \left( \frac{1}{R} + \frac{1}{r_1} + \frac{1}{r_2} \right) = 5 \left( \frac{1}{12} + \frac{1}{3} + \frac{1}{4} \right) Imax=5(1+4+312)=5(812)=4012 AI_{\text{max}} = 5 \left( \frac{1 + 4 + 3}{12} \right) = 5 \left( \frac{8}{12} \right) = \frac{40}{12}\ \text{A}

4. Ratio of Maximum to Minimum Current:

ImaxImin=4012512=405=8\frac{I_{\text{max}}}{I_{\text{min}}} = \frac{\frac{40}{12}}{\frac{5}{12}} = \frac{40}{5} = 8

Thus, the ratio of the maximum to the minimum current drawn from the battery source is 8.

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