A) 4.5 mH
B) 1 mH
C) 0.5 mH
D) 0.25 mH
Correct Answer: D
Solution :
Here, e.m.f. of primary coil \[{{\varepsilon }_{p}}=2mV=2\times {{10}^{-3}}V\] e.m.f. of secondary coil \[{{\varepsilon }_{s}}=5mV=5\times {{10}^{-3}}V\] Rate of change of current = 20 amp/sec From the Faradays law the mutual induction of secondary coil is given by \[M=\frac{{{\varepsilon }_{s}}}{di/dt}=\frac{5\times {{10}^{-3}}}{20}\] \[=2.5\times {{10}^{-4}}H=0.25\times {{10}^{-3}}H\]
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