The network shown in the figure is part of a circuit. If at a certain instant, the current I is 5A flowing through the circuit part and it is decreasing at a rate of 103As−1 then VB−VA equals
A
20V
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B
15V
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C
10V
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D
5V
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Solution
The correct option is B15V Current I is decreasing, hence emf will be induced across inductor such that it supports the current or will increase the decreasing current. ⇒+ve terminal of the induced emf will be facing towards B.
Now, potential diffrenece across the inductor is given as, |ε|=Ldidt=(5mH)×(103As−1) |ε|=5V ⇒Travelling from A→B and potential will drop across resistance by amount iR, potential will increase on crossing battery and inductor. VA−iR+15+Ldidt=VB VA−(5×1)+15+5=VB ∴VB−VA=15V