ElectroMagnetic InductionHard
Question
When the current in the portion of the circuit shown in the figure is 2A and increasing at the rate of 1A/s, the measured potential difference Vab = 8V. However when the current is 2A and decreasing at the rate of 1A/s, the measured potential difference Vab = 4V. The values of R and L are :
Options
A.3 ohm and 2 Henry respectively
B.2 ohm and 3 Henry respectively
C.10 ohm and 6 Henry respectively
D.6 ohm and 1 Henry respectively
Solution
Vab = L 
+ I R
8 = L × 1 + 2 × R
4 = − L × 1 + 2 × R
Solving the above equations
we get R = 3Ω
L = 2H.
+ I R8 = L × 1 + 2 × R
4 = − L × 1 + 2 × R
Solving the above equations
we get R = 3Ω
L = 2H.
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