# Inductor and resistance circuit

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#### matthew187

##### Junior Member level 2
Hi people, I was wondering if someone can help me with this revision question although i'm studying level 3 engineering this is a level 4 question, I've attempted the first question:-(

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;

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

#### godfreyl

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

2. What is the current in the inductor after the switch is opened?
2.5A because the current through an inductor can't change instantaneously.

3. What is the current in the resistor after the switch is opened?
-2.5A because the current through the inductor has to go somewhere, and that resistor is the only available path.

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

5. Given this, what is the voltage across the inductor after the switch is opened?
-500V, same as the resistor. Duh!

6. Given this, what is the induced EMF in the inductor?
OK, this looks like a trick question. I think you have to take into account the 80Ω resistance of the coil, so EMF = -2.5A * (200Ω + 80Ω) = -700V

tpetar

### tpetar

Points: 2

#### matthew187

##### Junior Member level 2
Silly of me i typed the wrong questions:???:

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
b) The current through the coil
c) The e.m.f induced in the coil
d) The voltage across the coil

I need to work out for each one immediately before the switch is open and immediately after the switch is open.

#### FvM

##### Super Moderator
Staff member

Just out of curiosity, if level 4 engineering brings simple RL problems, how many levels do you need until you fully understand electric circuits?

tpetar

### tpetar

Points: 2

#### matthew187

##### Junior Member level 2
Its a unit in mechanical engineering called engineering science, so I am doing level 3 in mechanical engineering, where if I was doing level 3 electronics engineering all this stuff would have been taught in a greater depth. Anyway how was question 3, -2.5a how did it become negative?

#### FvM

##### Super Moderator
Staff member
so I am doing level 3 in mechanical engineering, where if I was doing level 3 electronics engineering all this stuff would have been taught in a greater depth.
Thanks!
Anyway how was question 3, -2.5a how did it become negative?
The basic point, already mentioned by godfreyl is continuity of inductor current. It can only change continuously with finite voltage across the inductor. So IL1 (before opening the switch) must be equal to IL2 (immediately after opening the switch). IL1 = IL2 = 2.5 A. Everything else can be derived from this fact by DC network laws.

tpetar

### tpetar

Points: 2

A picture....

#### matthew187

##### Junior Member level 2
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
?
Immediately after Switch opening
-2.5a *(200Ω + 80Ω) = -700v

e) The voltage across the coil
Immediately before Switch opening
?
Immediately after Switch opening
-2.5a * 200ohms = -500v

So from what I have been explained and what I have understood I've gave these questions ago I've used ohms law and not no fancy exponential formula as I'm not confident on using them how far am I of achieving what I want to achieve from these questions? The bits I couldn't answer and don't understand i have marked with a "question mark".

#### FvM

##### Super Moderator
Staff member
a) The current Through the resistor
Immediately before Switch opening
200v/ 80ohm =2.5a
Immediately after Switch opening
200v / 80ohms = 2.5a
You meaned to say the current through resistor R2.
The bits I couldn't answer and don't understand i have marked with a "question mark".
All numbers in question can be read from the drawing.
e.m.f is zero while current is constant (with switch closed)
Voltage across the coil is equal to supply voltage. (It must be when the switch is closed).
I've used ohms law and no fancy exponential formula
Right. Exponential formula would be needed to calculate e.g. the current after 1 ms.

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#### matthew187

##### Junior Member level 2
a) The current Through the resistor
Immediately before Switch opening
200v/ 80ohm =2.5a
Immediately after Switch opening
200v / 80ohms = 2.5a

Yes this above question is refering to R2.

So if EMF is 0v then.....
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

So does this calculation look correct now?

#### FvM

##### Super Moderator
Staff member
I think it's correct.

#### matthew187

##### Junior Member level 2
so 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/ 80Ω=2.5a
Immediately after Switch opening
200v / 200Ω= 1a

b) The current through the coil
Immediately before Switch opening
200v / 80Ω = 2.5a
Immediately after Switch opening
200v / 80Ω = 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 * 80Ω = 200v
Immediately after Switch opening
-2.5a * 200Ω = -500v

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#### godfreyl

so all this is correct now?
No. I think you misunderstood earlier. In the picture, R1 is the resistor referred to in the question. R2 represents the resistance of the coil.

a) The current Through the resistor
Immediately before Switch opening
1A
Immediately after Switch opening
-2.5A

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