Hi Mr Al and WTP pepper, how are you? I've went away and this is what I've come up with I'm a bit more confident than before. As soon as this question is complete so is the assignment and I will surely post it on here the full assignment just so you can see what you guys have helped me achieve and if I get top marks which is a merit you guys get the credit
A coil of inductance 4H and resistance of 80Ω is connected in parallel with a 200Ω of resistor of negligible inductance across a 200v dc supply. The switch connecting these to the supply is then opened; the coil and resistor remain connected together. State in each case one for immediately before and one for immediately after opening the switch;
a) The current Through the resistor
Immediately before Switch opening
200v/ 80ohm =2.5a
Immediately after Switch opening
200v / 80ohms = 2.5a
b) The current through the coil
Immediately before Switch opening
200v / 80ohm = 2.5a
Immediately after Switch opening
200v / 80 ohm = 2.5a
c) The e.m.f induced in the coil
Immediately before Switch opening
2.5a * 80ohms -(200v) = 0v
Immediately after Switch opening
-2.5a *(200Ω + 80Ω) = -700v
d) The voltage across the coil
Immediately before Switch opening
2.5a * 80 ohm = 200v
Immediately after Switch opening
-2.5a * 200ohms = -500v
Anyway like I said I'm pretty confident on this one but again not 100% lol. So if you do spot something please do tell me.
A coil of inductance 4H and resistance of 80Ω is connected in parallel with a 200Ω of resistor of negligible inductance across a 200v dc supply. The switch connecting these to the supply is then opened; the coil and resistor remain connected together. State in each case one for immediately before and one for immediately after opening the switch;
a) The current Through the resistor
Immediately before Switch opening
200v/ 80ohm =2.5a
Immediately after Switch opening
200v / 80ohms = 2.5a
b) The current through the coil
Immediately before Switch opening
200v / 80ohm = 2.5a
Immediately after Switch opening
200v / 80 ohm = 2.5a
c) The e.m.f induced in the coil
Immediately before Switch opening
2.5a * 80ohms -(200v) = 0v
Immediately after Switch opening
-2.5a *(200Ω + 80Ω) = -700v
d) The voltage across the coil
Immediately before Switch opening
2.5a * 80 ohm = 200v
Immediately after Switch opening
-2.5a * 200ohms = -500v
Anyway like I said I'm pretty confident on this one but again not 100% lol. So if you do spot something please do tell me.