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Question:

An inductor of inductance L=400mH and resistors of resistances R1=2Ω and R2=2Ω are connected to a battery of emf 12V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at t=0. The potential drop across L as a function of time is :

6e⁻ᵗ/⁰·²V

12te⁻ᵗ/⁰·²V

12e⁻ᵗ/⁰·²V

6 (1−e⁻ᵗ/⁰·²)V

Solution:

I=ε/R=12/2=6A
ε=L(dI/dt)+R2I2
I2=I0(1−e⁻ᵗ/tc) ⇒I0=ε/R2=12/2=6A
tc=L/R=400×10⁻³=0.2
I2=6(1−e⁻ᵗ/⁰·²)
Potential drop across L=ε−R2I2=12−2×6(1−e⁻ᵗ/⁰·²)=12e⁻ᵗ/⁰·²V