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Circuit Problem - can you check my answers

sunny1982

Hi can someone just please check this for me and see if its correct, if not can you please give me a pointer to where I am going wrong 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

Reply 1

Stonebridge

Well if the switch is where the text seems to imply it is...
... if the switch is open...
Why would there be any current in any of the components?
Why would there be any pd across any of the components?

If the current is steady or zero there will be no emf induced in the coil.

If there is a pd across any component then all 3 will have the same value if they are in parallel. Yes?

Check your answers again.

I assume this is the complete question exactly as in your book/exam paper?
Was there a diagram?
As it stands there seems little point in the coil's inductance being given unless there is more to this question than you have given

Reply 2

sunny1982

No there isn't a diagram just the question, I attempted wouldn't there be 700v in the opposite direction?

Reply 3

Stonebridge

Original post by sunny1982

...wouldn't there be 700v in the opposite direction?

Why? Where does the 700V come from?
This is stated as a DC supply.
There would be a back emf generated in the coil at the moment the switch is closed and opened when the current rapidly increases to its steady value and decreases to zero, but there is no way of knowing what this is from the information given.
Anyway,the question refers to the situation before and after the switch is opened or closed. Not during.

Reply 4

sunny1982

So how would I tackle this problem its revision notes I am doing for an exam, I need to understand this just incase anything complex like this pops up in the exam.

Reply 5

Stonebridge

Firstly I would keep to "official" questions either from the exam board or a good text book.
This question is not very well constructed. Is this the complete question and with the exact wording?

If the circuit is DC (not AC) then the inductor acts as if it just had the 80 ohm resistance specified. So you just have two resistors in parallel connected to the supply.
If the supply was AC then things would be different.

Reply 6

sunny1982

The circuit is DC, and the question is exactly worded but I know it can't be complexed it was set by a college lecturer its got to be a simple solution to this :frown:

Reply 7

Stonebridge

Original post by sunny1982

The circuit is DC, and the question is exactly worded but I know it can't be complexed it was set by a college lecturer its got to be a simple solution to this :frown:

It is a simple solution as I pointed out in my earlier posts.

As there is no circuit diagram it seems you have a cell a resistor and a coil all in parallel.

With the switch open there is no current in and no pd across any of the components. (The questions says the switch "connecting these to the supply")

With the switch closed you just have a basic problem with two resistors (80 and 100 ohms) connected in parallel to a 200V supply.
Your answers for the current in both with switch closed are correct.
With switched closed all have the same (supply) pd.
At a steady DC current there is no induced emf in the coil.

Reply 8

uberteknik

Original post by sunny1982

The circuit is DC, and the question is exactly worded but I know it can't be complexed it was set by a college lecturer its got to be a simple solution to this :frown:

Seems to be the words 'immediately before' and 'immediately after' are the key. This is a clue to what the answer is looking for. i.e. the energy stored by the inductor (1/2 LI²) and what happens to the voltages and currents during the initial, transient and steady state conditions of the open and closed switch when that energy is released.

In other words, the question is looking for the step response for that circuit.

The inductor is 4H (this is a hefty coil with a lot of windings) with a winding resistance of 80 ohms (logically fits with the number of windings to make up 4H) is in parallel with a 200 ohm resistor and both connected to the 200v d.c. supply.

With the switch closed we can assume that immediately before opening, everything is stable. So.....

SWITCH CLOSED (Initial conditions):

The current flowing through the inductor will be 200v/80ohms = 2.5A (because it will behave simply like a resistor with a steady current flowing through it).
The current flowing through the parallel resistor will be 200v/200ohms = 1A
Total supply current will be 2.5A + 1A = 3.5A. because the current paths are parallel.

The voltage across both the inductor and the the 200 ohm resistor will simply be the supply voltage = 200V d.c.

The e.m.f. generated across the inductor will always oppose any change in the incoming current and is given by v(t) = L di/dt. (Faraday and Lenz's laws).

But the rate of change of current is zero. (it's steady state d.c. with the switch closed and therefore constant). So L di/dt = 0 hence the e.m.f. across the coil is 0.

SWITCH OPENS (transient condition)

We need to assume that the time taken to open the switch is abrupt and infinitely small.

The voltage from the supply is now zero so now no current is supplied from the d.c. supply.

However the inductor stored energy still remains and must now be dissipated in the 200 ohm resistor which forms a load since it is in parallel with the inductor. In essence the inductor becomes the limited power supply to the 200 ohms load resistor.

At the instant the switch is opened, the magnetic field will start collapsing but will try to maintain itself by generating a back e.m.f. of infinitely large size. i.e. because (Faraday and Lenz) it tries to oppose the change in loss of supply current which has collapsed to nothing in zero time.
L di/dt = -ve infinity)

But, there is a 200 ohm resistor connected in parallel which limits both the current and collapses the back e.m.f. very rapidly

Therefore at the infinitely small instant the switch is opened, the voltage across the resistor will try to achieve -ve infinity but will be limited by the parallel 200 ohns resistor.

The voltage across the resistor for practical purposes is therefore:

-2.5A x 200 ohms = -ve 500V (initial current reverse sign since it must now flow out of the inductor and from ohms law V = IR)

The maximum current through the 200 ohm resistor must therefore be -ve 2.5A. The voltage across the resistor will be -ve 500V rising to zero following an exponential law governed by the time constant t=L/R; 4/200 = 20 ms.

This is an exponential decay with a time constant given by t = L/R. (note the similarity to the capacitor/resistor time constant C.R).

i.e. after 20ms the voltage will have risen to 63.2% of -500V = -316V

(edited 11 years ago)

Reply 9

uberteknik

Be careful of the sign conventions. Remember an inductor stores magnetic energy which is not the same as the charge storage of a capacitor.

Reply 10

Stonebridge

Yes, apologies to sunny 1982 for my misinterpreting the words "immediately after" opening the switch.
Thanks uber for spotting this.

Reply 11

sunny1982

Guys this is brilliant fabulous help, I couldn't thank you enough its amazing the way you explain things uberteknik. I was wondering if you can explain this to me aswell please,

How does current and voltage vary with time in series RC and RL circuits both on connecting a supply and shorting out the circuits?

Reply 12

uberteknik

Original post by sunny1982

I'm assuming you mean a series RC and series RL connected to a d.c. supply? But what do you mean by 'shorting out' the circuits?

1) RC. The initial state will be no charge on the capacitor so there will be no potential voltage developed across its terminals.

The current through the capacitor is given by I(t) = dv/dt. So with t = 0 the capacitor looks like a short circuit to the supply.

The current will, however, be limited by the series resistor to Vsupply/R intially, but as the charge on the plates starts to build up, a voltage, V=Q/C, across the capacitor will accumulate.

But the supply voltage has not changed so the voltage across the series resistor will immediatey start to fall; Vresistor = Vsupply - Vcapacitor.

Hence the current availabe to charge the capacitor will also fall in response (I= Vres/R = (Vsupply - Vcapacitor)/R and the rate of charge slows as does the rate of rise of the voltage across the capacitor.

The voltage across the cap' follows an inverse exponential increase with a time constant given by t=CR reaching 63.2% of the supply voltage in that time.

The current into the capacitor thus follows an exponential fall from an initial value of I=Vsupply/R and decays to zero following the same t=CR time constant. i.e. it will fall to 63.2% of the initial current in that time.

(edited 11 years ago)

Reply 13

sunny1982

Ok lets leave the bit about shorting out the circuit so what happens to an RL circuit?

Reply 14

uberteknik

Original post by sunny1982

Ok lets leave the bit about shorting out the circuit so what happens to an RL circuit?

Capacitors store energy in the electrostatic field between the charge plates. Inductors store energy in the magnetic field between the coil windngs.

So, you have a go. What do you think is happening?

Reply 15

sunny1982

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

Anyway uberteknik heres all them questions answered and because of your expert help I am quiet confident when I answered these the second time lol