# Homework Help: Coils and inductance

1. Apr 9, 2013

### Andrew187

Hi there I needed someone to check my answers for these questions and see if they are correct and if not if they can point me in the right direction the question is for revision purposes got an exam in a week.

Thanks.

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;

The current Through the resistor

Switch closed
200v/200Ω =1a
Switch open
200v/80Ω =2.5a

The current through the coil

Switch Closed
200v/80Ω = 2.5a
Switch open
200v/(200ohm +80Ω) = 0.714a

The e.m.f induced in the coil

Switch closed
(2.5a*80Ω)/200v = 0v
Switch Open
2.5a *(200Ω + 80Ω) = 700v

The voltage across the coil

Switch closed
200v/1a = 200v
Switch open
2.5a * 200Ω = 500v

2. Apr 9, 2013

### phyzguy

These aren't right. You have the 'switch closed' numbers right, but not the 'switch opened' ones. Immediately after the switch is opened, the current through the inductor will be the same as immediately before the switch is opened, since the current through the inductor cannot change instantaneously. All of this current then goes through the 200 ohm resistor, since it has nowhere else to go. Try using this way of thinking to calculate the other quantities.

3. Apr 9, 2013

### Andrew187

Thats where I'm having the problem, I don't seem to understand how to tackle this problems or what formula's to use this unit was covered last year and even though I'm going through my notes I can't seem to tackle the problem.

4. Apr 9, 2013

### phyzguy

Try drawing the circuit with the currents immediately before and after the switch is opened.

5. Apr 9, 2013

### Andrew187

Why would I need to draw the circuit? I think i'm making the question more complexed than it sounds I just need to know how to calculate them.

6. Apr 9, 2013

### phyzguy

Well, try doing what I told you in Post #2, and answer the following questions, in order:

(1) What is the current in the inductor before the switch is opened?
(2) What is the current in the inductor after the switch is opened?
(3) What is the current in the resistor after the switch is opened?
(4) What is the voltage across the resistor after the switch is opened?
(5) Given this, what is the voltage across the inductor after the switch is opened?
(6) Given this, what is the induced EMF in the inductor?

7. Apr 9, 2013

### Andrew187

(1) 200v/ 80ohms = 2.5a
(2) 200v/200ohms = 1a
(3)
(4) 2.5 * 80 =200v
(5)
(6)
I just had a feeling it would be wrong the questions I have answered I answered them confidently rest of them have been a stab in the dark.

8. Apr 9, 2013

### phyzguy

Correct.
No! The current through the inductor cannot change instantaneously. Since V = L dI/dt, if the current changed instantaneously, dI/dt would be infinite, which would require an infinite voltage. So the current in the inductor after the switch is opened is the same as before the switch is opened, 2.5A.

Using this, try answering questions 3-6.

9. Apr 10, 2013

### Andrew187

(2) 200v / 80 =2.5a
(3) 200v/200 ohms = 1a

Is this correct?

10. Apr 10, 2013

### phyzguy

No, After the switch is opened, the 2.5A flowing through the inductor must flow through the resistor, because it has nowhere else to go. Kirchoff's current laws tell you this. So if 2.5A is flowing through the 200 ohm resistor, what is the voltage across this resistor?

11. Apr 10, 2013

### Andrew187

What is the current in the inductor before the switch is opened?
200v/80hms= 2.5a

What is the current in the inductor after the switch is opened?
200v/ 80ohms= 2.5A

What is the current in the resistor after the switch is opened?
200v/-80 = -2.5A

What is the voltage across the resistor after the switch is opened?
-2.5A * 200ohms = -500V

Given this, what is the voltage across the inductor after the switch is opened?
-2.5* 200-500V

Given this, what is the induced EMF in the inductor?
-2.5A * (200ohm + 80ohm) = -700V

12. Apr 10, 2013

### phyzguy

Great! I think this is now correct.

13. Apr 10, 2013

### Andrew187

Great so to be sure all this is correct now?

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/ 200ohm = 1a
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

Last edited: Apr 10, 2013
14. Apr 10, 2013

### phyzguy

The current through the resistor immediately before the switch is opened is 1.0 A, like you had in post #1. I think the rest is correct.