Two inductors \(L_{1}\) (inductance 1 mH, internal resistance 3 \(\Omega\)) and \(L_{2}\) (inductance 2 mH, internal resistance 4 \(\Omega\)), and a resistor \(R\) (resistance 12 \(\Omega\)) are all connected in parallel across a 5 V battery. The circuit is switched on at time \(t=0\). The ratio of the maximum to the minimum current \((I_{\text{max}}/I_{\text{min}})\) drawn from the battery is

JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Electromagnetic Induction

Two inductors

L_{1}

(inductance 1 mH, internal resistance 3

Ω

) and

L_{2}

(inductance 2 mH, internal resistance 4

Ω

), and a resistor

R

(resistance 12

Ω

) are all connected in parallel across a 5 V battery. The circuit is switched on at time

t = 0

. The ratio of the maximum to the minimum current

(I_{max} / I_{min})

drawn from the battery is

Answer: 8

Solution

t = 0

, the inductors act as open circuits (infinite impedance), so current flows only through the resistor

R

I_{min} = \frac{V}{R} = \frac{5}{12}

A

At

t \to \infty

, the inductors act as short circuits (zero impedance), so current flows through all three parallel branches with only their resistances:

Equivalent resistance:

R_{eff} = (\frac{1}{3} + \frac{1}{4} + \frac{1}{12})^{- 1} = (\frac{4 + 3 + 1}{12})^{- 1} = \frac{12}{8} = \frac{3}{2}

Ω

I_{max} = \frac{V}{R _{eff}} = \frac{5}{3/2} = \frac{10}{3}

\frac{I _{max}}{I _{min}} = \frac{10/3}{5/12} = \frac{10 \times 12}{3 \times 5} = 8

JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Integer answer

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JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Electromagnetic Induction

Solution

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