Electromagnetic Induction

JEE Main level, numerical answer, medium

147 of 153

Question 147 of 153

As per the given figure, if dIdt=−1As−1\dfrac{dI}{dt}=-1A{s}^{-1}, then the value of VAB{V}_{AB} at this instant will be ______ VV.

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Answer: 30


Given here,
dIdt=−1As−1\dfrac{dI}{dt}=-1A{s}^{-1}
Applying Kirchhoff's voltage law to the circuit,

VA−IR−LdIdt−12=VB{V}_{A}-IR-L\dfrac{dI}{dt}-12={V}_{B}
⇒VA−2×12−6(−1)−12=VB\Rightarrow {V}_{A}-2\times 12-6(-1)-12={V}_{B}
⇒VAB=VA−VB=36−6=30V\Rightarrow {V}_{AB}={V}_{A}-{V}_{B}=36-6=30V

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