A circuit that contains three identical resistors with resistance `R = 9.0 Omega` each, two identical inductors with inductance L = 2.0 mH each, and an ideal battery with emf `epsilon = 18 V` . The current 'i' through the battery just after the switch closed

Explanation

Because the current through each inductor is zero before the switch is closed, it will also be zero just afterward. Thus, immediately after the switch is closed, the inductors act as broken wires, as indicated in Fig. 30-31b. We then have a single-loop circuit for which the loop rule gives us
`E-iR=0`
Substituting given data, we find that
`i=E/R=(18V)/(9.0Omega)=2.0A`
We now have a circuit with three identical resistors in parallel, from Eq. 27-23, their equivalent resistance is `R_(eq)=R//3=(9.0Omega)//3=3.0Omega`. The equivalent circuit shown in Fig. 30-31d then yields the loop equation `E - iR_(eq) = 0`, or
`i=E/(R_(eq))=(18V)/(3.0Omega)=6.0A`