# Inductors and resistance

Discussion in 'Electronics Homework Help' started by matthew187, Apr 10, 2013.

1. ### matthew187

5
0
Apr 10, 2013
Hi I was wondering if some on could check these answers and see if I am along the right lines and if someone could be very kind and help me with the bits that are "question marked" as I don't understand these its just a revision question. many 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;

a) The current Through the resistor
Immediately before switch opening
200v/ 80ohm =2.5
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 / 80ohm = 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

2. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,482
2,830
Jan 21, 2010
Remember that the resistor has a resistance of 200 ohms.

Where does the current across that resistor come from once the switch is opened?

This looks right

What induces this EMF?

Is this factor present with the switch closed?

Draw the circuit. The answer should be obvious.

3. ### matthew187

5
0
Apr 10, 2013
What I took into consideration is that the 200 ohm resistor is negligible inductance, so I've took into consideration what you have said and this is what I've come up with.

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

4. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,482
2,830
Jan 21, 2010
Draw the circuit and look at it.

Q1 What is the value of the resistor? (hint, it's not 80 ohms)

Hint2: part 1 of Q1 is wrong, but part 2 is right except for the sign.

5. ### matthew187

5
0
Apr 10, 2013
I have tried to draw the circuits, but what I don't understand is that 200ohm is neglible inductance so how does that come into play?

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6. ### davennModerator

13,802
1,941
Sep 5, 2009
200 Ohm is neglible inductance = thats irrelevent look at it for what it is, a resistive load across the supply in parallel with the resistance of the inductor.....

Dave

7. ### matthew187

5
0
Apr 10, 2013
I've gone away thought about it and this is what I've come up with.

a) The current Through the resistor
Immediately before Switch opening
200v/ 200Ω= 1a
Immediately after Switch opening
-200v / 80Ω= -2.5a

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

Last edited: Apr 11, 2013
8. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,482
2,830
Jan 21, 2010
The answers look pretty good to me.

I'm not sure you always chose the easiest way to get there, but the numbers seem right.

9. ### matthew187

5
0
Apr 10, 2013
I've used ohms law, I didn't think I needed any fancy formulas for this like L di/dt but I don't know how to use them