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Inductors and resistance

matthew187

Apr 10, 2013
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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
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Jan 21, 2010
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a) The current Through the resistor
Immediately before switch opening
200v/ 80ohm =2.5
Immediately after Switch opening
200v / 80ohms = 2.5a

Remember that the resistor has a resistance of 200 ohms.

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

b) The current through the coil
Immediately before Switch opening
200v / 80ohm = 2.5a
Immediately after Switch opening
200v / 80ohm = 2.5a

This looks right

c)The e.m.f induced in the coil
Immediately before Switch opening
?
Immediately after Switch opening
-2.5a *(200Ω + 80Ω) = -700v

What induces this EMF?

Is this factor present with the switch closed?

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

Draw the circuit. The answer should be obvious.
 

matthew187

Apr 10, 2013
5
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Apr 10, 2013
Messages
5
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
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
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25,505
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.
 

matthew187

Apr 10, 2013
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Apr 10, 2013
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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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davenn

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Sep 5, 2009
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I have tried to draw the circuits, but what I don't understand is that 200 Ohm is neglible inductance so how does that come into play?

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
 

matthew187

Apr 10, 2013
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Apr 10, 2013
Messages
5
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:

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
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Messages
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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.
 

matthew187

Apr 10, 2013
5
Joined
Apr 10, 2013
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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.

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 :(
 
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