# Electric circuits

1. Jan 10, 2014

### Coco12

Problem statement
What would the voltmeter on the parallel circurt read? In the diagram.

Revelant equations :
V=IR

Attempt at solution:

Just wanted to confirm a fact : the 10 V given in the circurt drawing would be the drop in potential meaning that it lost 10 V so the potential energy moving out of the resistor would be 8, right? Or would it meant that it dropped from 18 to 10?

View attachment 65497

2. Jan 10, 2014

### Coco12

3. Jan 10, 2014

### tiny-tim

Hi Coco12!

(I've never seen a diagram like this before … when you buy a resistor in the shop, it's marked in ohms, not in volts!! but anyway …)

Yes, i think you're correct …

the clue is that they add up to 18 …

each marked voltage is the voltage drop across that component, ie voltage in minus voltage out.

4. Jan 10, 2014

### Coco12

5. Jan 10, 2014

### tiny-tim

Hi Coco12!
current moves through things …

the current going in is the same as the current coming out​

voltage doesn't move through things, voltage just is

(voltage is electric potential, it's just like gravitational potential per mass)

10V is the difference in electric potential (voltage) between the first strech of wire and the second

(why 5 ? )

the voltmeter is attached between the first two resistors, and between the third resistor and the battery

it measures the electric potential difference (the voltage drop) between those two points

6. Jan 10, 2014

### Coco12

The second stretch of wire has a voltage of 8 then. Between those two points, the voltmeter would read 8.

7. Jan 10, 2014

### tiny-tim

yes, the voltmeter is attached at 8V above the level of the battery, and at 0V, difference = 8V

(it would be better if you said the second stretch of wire has "a difference in voltage", or "a voltage drop")

8. Jan 10, 2014

### Coco12

Thank you tiny tim I understand it now. I was thinking that the voltage would separate as there is an additional pathway for the circurt therefore the voltmeter would read 3 A but I guess that would only apply to current splitting up and reconvening. Voltmeter is used to calculate the difference in potential energy not the amount flowing through it right?

9. Jan 10, 2014

### tiny-tim

the current separates at a junction, the voltage doesn't

the voltage drop across the two resistors is exactly the same as the voltage drop across the voltmeter (the voltage difference between two points is independent of the path taken, just as the gravitational potential difference is independent of the path taken)

(a voltmeter usually has a very high resistance, so that nearly all the current continues through the original circuit, and only a tiny current goes through the voltmeter itself

technically, the resistance of the voltmeter is in parallel to the two resistors, so it will decrease the total resistance slightly, which will mean a slightly higher voltage drop through the first resistor, and a slightly lower one through the other two … but the error is less than the accuracy of the voltmeter itself)
yes