1. The problem statement, all variables and given/known data For the circuit shown in the figure, the switch has been open for a very long time. (a) What is the potential drop across the 15.0-mH inductor just after closing the switch? (b) What is the potential drop across the 70.0-µF capacitor after the switch has been closed for a very long time? 2. Relevant equations I = V/R (1 -e-Rt/L) V = IR = Q/C 3. The attempt at a solution (a) I replaced the capacitors with wires at t=0 and found the current in the outer loop from V = IR 200V = 100ohm *I I = 2A Then the resistance through the loop with the inductor is 75ohm, so V = 2A *75ohm = 150V (b) I replaced the inductors with wires from the limit at t=infinity of the equation for current through the inductor and the capacitors with an open circuit, so there is no current through the capacitors, and the current through the middle loop is 200V = I *75ohm I = 2.6... A I'm confused if there is no current through the loop with the 70-µF capacitor, what resistance is used to find the voltage? The voltage is not 0 or 200V (the battery supply). I also know the three branches are technically in parallel.