Question
GeneralGeneralGeneral

An inductor of inductance 10 mH having resistance 100 Ω is connected to a battery of EMF 1.0 V through a switch as shown in the figure.

After the switch is closed, find the ratio of instantaneous voltages across the inductor when the current through it is:

  • I₁ = 2 mA
  • I₂ = 4 mA

Options:

  • (1) 3/5
  • (2) 3/4
  • (3) 4/3
  • (4) 5/3

Verified Answer

In an RL circuit, Kirchhoff's voltage law gives:

E = IR + VL

where:

  • E = 1 V
  • R = 100 Ω
  • VL = voltage across inductor

For current I₁ = 2 mA:

VL1 = 1 − (0.002 × 100)

= 1 − 0.2

= 0.8 V

For current I₂ = 4 mA:

VL2 = 1 − (0.004 × 100)

= 1 − 0.4

= 0.6 V

Required ratio:

VL1 / VL2 = 0.8 / 0.6 = 4/3

Final Answer

Option (3) 4/3