# Circuit Analysis help for a simple circuit

Discussion in 'Homework Help' started by rakeshkool27, Jan 7, 2011.

1. ### rakeshkool27 Thread Starter New Member

Jan 7, 2011
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Hello friends,I was trying circuit analysis of a simple circuit but am stuck and confused with the results found by multimeter.The circuit diagram is given in the .jpg file.I got the value of I1 and I2 as 120mA and -113.5mA and hence the value of I3 as 6.5mA. But how do I get this value theoretically? I could neither apply nodal analysis,mesh anlaysis nor superposition theorem successfully.
If you could find some clue to it,please let me know.
Thanks.

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2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Circuit in Figure 1 can not be solve becaues we don't know internal resistance of a voltage source.

3. ### rakeshkool27 Thread Starter New Member

Jan 7, 2011
15
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Okay. So if there is a resistor available connected to E1 and E2 then only its possible to apply nodal or mesh analysis,otherwise such circuits are not readily possible.
And in the E2 branch the current flows in opposite direction.So it is due to the fact that E1>E2,right??

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Well in theory current will reach infinity, because zero resistance an there is a voltage between to batteries
For example if we assume internal resistance equal 0.01Ω
Then I1 = 150A I2= -150A ; I3 = 6.5mA.
So 5V battery will be charged via 8V with 150A of current

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5. ### Adjuster Well-Known Member

Dec 26, 2010
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Yes, that's right, a theoretical solution of Fig.1 would require knowledge of internal resistances, so that they could be included in the calculations.

A practical realisation of Fig. 1 could be very dangerous, The OP should avoid doing this again. Depending on the internal resistances of the batteries used, a very large current could circulate between them. With many types of battery, this could lead to a fire, or even an explosion. I am surprised that the currents measured were as low as 100mA - perhaps tired old batteries have saved the OP from having an accident.

Last edited: Jan 7, 2011
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6. ### rakeshkool27 Thread Starter New Member

Jan 7, 2011
15
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Well, I get the picture now...Thanks

7. ### rakeshkool27 Thread Starter New Member

Jan 7, 2011
15
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Actually I practically tried out and it was great shock hearing that.But thank god,there's been no such accident and the power supply is 5V and 8V adapter not a battery.I think thats the reason.

8. ### rakeshkool27 Thread Starter New Member

Jan 7, 2011
15
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what can be said about the 2nd circuit...Its an OR gate.
This circuit is also similar to the 1st circuit,except the diodes.Isn't it also dangerous,since here also there is no resistance in the branches and only a 0.7V voltage drop will be there across the diode.
Suppose we give E1=10V and E2=0V,then it becomes a simple resistive circuit with a voltage drop of Vr=10V and current I1=.01A across 1KΩ resistor.Again if we give both E1 and E2 10V,then also Vr=10V still when the current I3 is doubled than before.
Why is it so??

And for the 1st circuit,again I would like to ask why cannot we consider the circuits separately for each source as we do in superposition theorem,and add the respective currents using KCL??

9. ### Adjuster Well-Known Member

Dec 26, 2010
2,147
301
You have not quite understood the second circuit. The diodes make a fundamental difference, because they prevent any significant current flowing into either of the batteries, provided that the difference between the two voltages is below the diode reverse breakdown voltage. This eliminates the possibility of currents circulating between the two supplies.

For the second circuit, if E1 and E2 are unequal, only the diode connected to the larger positive voltage will conduct. The other diode will be reverse biased and will not conduct. If E1 and E2 are equal, both diodes will conduct - with identical diodes each would then pass half of the resistor current. Note that I3 is not doubled in this case compared to when one of the two voltages is zero. You cannot employ the superposition theorem here, because diodes are non-linear devices: they violate the conditions required for superposition.

For the first circuit, superposition etc. could give the correct result if correctly applied to a circuit using batteries, but this would require their internal resistances to be taken into consideration.

Note however that the behaviour of a circuit like the first example but using mains power supplies is likely to be more complicated. A mains supply cannot always be represented as just a simple voltage source and a linear resistance, especially where reverse current flow is concerned. Here again, superposition may not give the expected results because of non-linear behaviour.

Putting two different small mains supplies in parallel may not result in such large current flows as might result from doing the same with some batteries, but there is still a real risk of fire or doing damage. There is further danger because mains voltages are present.

The power supplies which have been used in this way may therefore have been damaged, and may already have failed or be subject to failure in future. You should not use these supplies for any other purpose unless they can be proven to be in safe working order.

Last edited: Jan 8, 2011
10. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,173
1,188
No, this circuit is safe thanks to diode OR gate.
In this circuit there is no way to current to flow between batteries.
For E1 = 8V and E2 = 5V
So D1 is ON and D2 is OFF.
And
Ie = (E1 - Vd) / R

Because E1 and E2 provide half of a load current

In theory it is impassible to have different voltages on element connect in parallel

11. ### raybo Member

Oct 18, 2008
22
0
there is only one circuit 8 volts to the resistor the 5 volt is biased off now figure it the amps. dont. forget 8 - .7v