# Circuit -Question No. 3

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#### randolfo

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Could you please help me how to approach the problem on this circuit? I am trying to solve how many voltage are there between the R2 and the ground ? How many amperes between R2 and the ground ? Perhaps if you could give me idea or better yet give me the solution and explanation how you arrived with the answer.

Thank you. Randolfo

Your scematic is hard to read. Try upload a bigger (original if possible) image, not a schrimped one.

The quality of the schematic image is too poor to see the component values. Try attaching a better quality image.

Sounds like a homework problem. How about making a first guess at what you think the answers are.

Hint: The zener diodes look like a constant voltage across their respective resistors.

rE: Circuit Ques 3

Sorry, for a small image posted previously. I hope this is enough. I don't how how would I approach the problem. I know zener diode will keep the voltage drop constant in the circuit. I am guessing 25 V presence at R2 and 2.5 A ? I don't know i hope you could help me with a clear explanation. I'm still a novice in circuit analysis, but i love learning it. . Thank you.

supply voltage

The supply voltage is 30 V . . how?

Could be a trick question - if they are applying a positive voltage on the cathode of D3, no current will flow and the nodes will all be zero volts.

No voltage at all because D3 doesn't conduct current.

No Voltage?

Yeah maybe right. No voltage because diode D3 doesnt conduct from anode to cathode? thats what I read from my manual . But are there others with confirmatory opinion to that answer? What about if the diode D3 is reversed , what will be the answers to two previous question? I am thinking I drawn the circuit wrong, I have to take a look again and make the changes if ever...

Well, if we assume that D3 is reversed and has a voltage drop of a 0.7V.
And the supply voltage is 30V.
So, from KVL we can write:

30V=VR3+VD2+VD1+VD3
VR3=30V-VD2-VD1-VD3=19.3V

And IR3=VR3/R3

I just checked again the given circuit, there is minor changes but almost the same. And I was given 4 choices . For the voltage between R2 and letter A the choices are 25V, -25 V, 6 V, -6V. For the current in R2 and letter A the choices are 2.5mA, -2.5mA, 6mA, -6mA. So i don't know how to come up with any of those answers . Maybe there is power supply somewhere in the circuit but it was not given because it was assumed to be there. I'm still confuse... Thank you.

hi,
I wont give the answer, but a method. Get the total resistance of the 3 resistors, calculate the current flowing thru the resistors.

Calculate the voltage drop across R1.

If its greater than 5V [D1 is 5V] then D1 is conducting, so there is 5V across D1.

The rest is up to you.

D1/D2 will clamp the voltage across R1/R2, respectively. and R3 takes the rest of the voltage.

the first one you posted will have the diode reverse biased so it will take all of the 30v, unless that exceeds its breakdown voltage.

It could be -30V.

I would imagine (hope?) that if it were -30v, the poster would have said -30v, not 30v.

okay the diagram is still a little confusing, there are two different ground symbols, at least thats all I learned those two symbols as. And the white triangle on the right side is labeled as "D3" like a diode, but its not a diode...

I will assume that the Ground symbol on the left is 0V reference and that the white triangle on the right is +30V (this is probably the 'power source' you mentioned).

ericgibbs - theres a much quicker solution to this problem than calculating currents in each resistor. I think the point the problem is trying to drive home is the effect of diodes.

Each diode is noted with a voltage. This is probably the intrinsic voltage drop of the diode. The ideal diode model states that if a diode has 0V or less across it, it does not conduct, and that if there is a positive voltage across it, it will conduct any current. Instead of 0V, your diagram says this voltage is 5V. So the diagram seems to instruct that every diode IS conducting, and each one has a 5V drop across it. (decent ideal diode explanation here: Diodes )

So you might first look at this and think: what do circuit elements in parallel have in common? Parallel circuit elements will have the same voltage. Series circuit elements will have the same current. Starting at the +30V reference (on the right) youll notice that there is a parallel combination of a resistor and diode. The diode is conducting, and it drops 5V. This means that the parallel combination of that diode (D1) and that resistor (R1) both have the same voltage across them: 5V.

Similarly, there is another, similar parallel combination to the left of that, again, with the 5V diode "clamping" the voltage across the resistor.

Now, you know the voltage across both R1 and R2. So now you know a voltage and a resistance, and are asked for a current (current in R2). You must know the circuit law to find that.

This is kind of interesting as the current in R2 is not the current in R3. If there are 2 5V diodes, that makes the voltage across R3 20V. 20V / 1kohm = 20 mA. So when you find the current in R2, you can also find the current in D2, and then the current in R1 and D1, and so on.

The point here is the "clamping" effect of the diodes.

Sorry if I solved too much, try another similar problem just to be sure youve grasped the concept.

• ericgibbs
hi solis,

I think when you refer to 'diodes' you mean zeners.?

Your method sounds far more convoluted than mine. wat is the name of this circuit??

wat is the name of this circuit??

Its only a circuit for testing students. hi solis,

I think when you refer to 'diodes' you mean zeners.?

Your method sounds far more convoluted than mine. oh, yes, they are zeners but everything still holds in my explanation, theyre just conducting in breakdown. oops.

as for it being convoluted, perhaps for a novice, but thats the point of the problem: learning (zener) diodes, not ohms law or KVL.

once you know how zeners work you can figure the voltage at any point in that circuit just by looking at it, and the current through any of the resistors almost as fast. though it takes far more text to describe an intuition-based solution.

your method checks to see if the zeners are in breakdown mode, which is the correct way to do it. but once you know that they are, you have to re-calculate the current in the resistors, hence where my solution comes in.

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