A \[5\,\,cm\] long solenoid having \[10\,\,\Omega \] resistance and \[5\,\,mH\] inductance is joined to a \[10\,\,V\] battery. At steady state, the current through the solenoid (in ampere) will be
A)\[5\]
B)\[2\]
C) \[1\]
D) \[zero\]
Correct Answer:
C
Solution :
At steady state, the current through the solenoid \[i=\frac{E}{R}=\frac{10}{10}=1\,\,A\]