Ended up figuring out how to solve this problem. I just was not expecting certain parts of the results which lead me to believe I did it wrong. 1. The problem statement, all variables and given/known data Voltage of battery is 5V Resistors are 5 Ohms The solenoid has no internal resistance 1.Need to find V_2 before switch is opened 2.The current through the solenoid right after the switch has been opened. 3. The current through R_2 and R_1 right after the switch has been opened. 4. Voltage across R_1 right after the switch has been opened. 5. Voltage across R_1 if it was 5 Mohms right after the switch has been opened. 2. Relevant equations V=IR Kirchhoff's Loop Rule That a solenoid will want to resist any change in current. 3. The attempt at a solution 1. The voltage across R_2 will be 5V because the loop that consists of the switch and resistor has to have a voltage drop of 5 V and the resistor is the only element with a voltage drop on that loop. 2. The current through the solenoid right after the switch is open will be 1 A as the solenoid will want to keep the same current as before. 3. 1 A as the current will be same as the solenoid. 4. The voltage drop across both of the resistors will be 5 V. The solenoid will add additional voltage to create the greater voltage. 5. 5*10^6 V as the current will still be 1 A. I know I probably have flawed logic in here somewhere though I just do not know exactly how a solenoid will effect this circuit after the switch is thrown.