# Question about potential difference in circuit with resistance

1. Jun 28, 2012

### iampaul

Consider a simple circuit composed of a battery whose ends are connected by a wire. The potential difference of the battery is Vbatt

ohm's law gives: V=IR
If the resistance of the circuit is ignored then the potential drop across the wire is zero.
But using kirchoff's voltage law the sum of the potential difference in the loop must be zero,my first question is where does the potential drop?

If the resistance of the wire is not ignored then there is a voltage drop equal to the potential difference of the battery.Even when the resistance is small approaching zero the voltage drop is still equal to Vbatt. The limit approached by the voltage drop as the resistance approaches zero is still Vbatt but the voltage drop when the resistance equals zero is zero by using V=IR. My 2nd question is why is it not continuous? Shouldn't the voltage drop still be Vbatt

My third question is about a switch in parallel with a resistor and a battery. The resistance of the conducting wires is ignored. My professor said that if the switch is open then there is zero current passing through it and using V=IR the voltage across the switch is zero. Because it is in parallel with the resistor the potential difference across the resistor is also zero. but can't a switch be considered as a conducting material(air) with high resistance. The total resistance of the parallel combination of the switch and the resistor is approximately equal to just the resistance of the resistor by using the equation: (1/rs)+(1/r)=(1/rT) since the ratio (1/rs) almost vanishes because rs, the resistance of the switch is too big. But since there is a total resistance there must be a voltage across the resistor.

Last edited: Jun 28, 2012
2. Jun 28, 2012

### Staff: Mentor

It's not physically realistic to just ignore the resistance of the circuit. While the resistance of the wire may be neglected compared to other circuit elements that have resistance, when the wire is all there is you can't neglect it. And realize that the current will be high when the resistance is low, so IR can still be finite. And batteries themselves have internal resistance.

Again, taking the limit as R goes to zero is not realistic. And realize that IR can still be finite.

There is no current passing through it, but the voltage across it will not be zero.
Your reasoning is fine. An open switch in parallel with resistor is equivalent to the resistor alone. There will be a voltage across the switch.

3. Jun 30, 2012

### iampaul

I have another question. It is a voltage source(V) in series with a resistor(R1) and with the parallel combination of a diode and another resistor(R2). The image of the circuit is attached below. My professor discussed the same circuit and said that if the diode is reverse biased, it acts as an open switch and the voltage across R2 is equal to V. It is a lesson about clipper circuits. My book stated the same thing without explanation so i guess the reason must be simple. But i still don't get it.
My attempt for an explanation is as follows:
The diode becomes an open switch when reverse biased. An open switch has high resistance, and has a potential difference across it. This potential difference equals the potential drop across R2 because it is in parallel with the diode, VD=V2. Considering the loop containing V, R1 and the diode, by Kirchoff's voltage rule, V-V1-VD=0 or V-V1-V2=0. Now V=V2 if V1=0.

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4. Jun 30, 2012

### iampaul

I'll try to read my book again to see if this is really what it means to say? Sometimes the english really makes it difficult for me.

5. Jun 30, 2012

### Staff: Mentor

Sounds good to me.
V1 is the voltage across R1. Why are you setting it to zero? (Perhaps you mean: If you removed the resistor R1, then V = V2. Sure.)

6. Jun 30, 2012

### CWatters

I believe you missunderstood or your prof didn't explain it well. The first part of that is correct BUT the voltage across R2 will only be equal to V if R1 is zero (or very small compared to R2).

Let me try and describe the circuit..

Lets assume the V in your circuit is 10V and R1=R2=1KOhm

In your diagram the top end of the battery is shown as +ve. That means the diode will be forward biased = conducting. It's like a switch that it "ON". So replace the diode in the diagram with a wire!

That means the current flowing in R1 will be

I = (V-0)/R1
=10V/1000Ohms = 10mA

OK so far?

Now turn the diode around so it's "pointing" upwards.

Now the diode is reverse biased = non conducting. This is like a switch that is "OFF" or open circuit. So you can remove the diode from the diagram because it is doing nothing.

Now the current flowing in R1 is

I = V/(R1+R2)
= 10V/2000ohms
=5mA

The voltage on R2 is given by the "potential divider" rule..

V(r2) = V * R2/(R1+R2)

So the voltage on R2 is NOT = V

However if R1 is very small or zero then..

V(r2) = V * R2(0+R2)
= V * R2/R2
=V

and the voltage on R2 is equal to the battery voltage.

R1 might be very small if it is intended to represent the "internal" or "source resistance" of the battery. Real batteries are not "perfect". They have a small internal resistance due to the way they are made. This is very small for a car battery but quite large for some AA batteries especially when nearly empty.

7. Jun 30, 2012

### CWatters

When you understand the above...

Consider what would happen if you built the circuit as it was drawn and R1 was only 0.1ohm?

The diode is conducting so the current is given by

I = (V-0)/R1
=10/0.1
= 100 Amps !!

Most diodes would fail very quickly. So in most circuits R1 would not be small.

8. Jun 30, 2012

### iampaul

thanks
I went over my notes, my professor did use the voltage divider principle, sorry for the post.

9. Jun 30, 2012

### iampaul

Sorry to disturb you but i have another question.

What if there is a circuit consisting only of a 12v source and a 9v source
If the batteries are ideal and i use Kirchhoff's loop rule it seems to show that 12V=9V??
I think i should take their internal resistance into consideration.
Another question is when a 9v voltage source is connected in parallel with a silicon diode. KVl gives 9v=barrier potential=0.7V