# Electricity -- Effect of heating a thermistor in this circuit

1. Apr 1, 2017

### Kajan thana

1. The problem statement, all variables and given/known data
The thermistor is heated so that its resistance decreases. State and explain the effect this has on the voltmeter reading in the following positions. A–C

2. Relevant equations
V=IR

3. The attempt at a solution:
I will say that the voltage will increase because if the overall resistance will decrease therefore there is more current flow. As voltage is directly proportional to current, then the voltage will increase also.

^^ I have attached the circuit to this question.

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2. Apr 1, 2017

### Staff: Mentor

When you specify "A-C", are you asking about the potential difference across the upper 20 kΩ resistor?

3. Apr 1, 2017

### Staff: Mentor

I don't think your answer is right. The reisistors A-E and B-F are in parallel. If the 12V voltage source is ideal (zero internal resistance), what can you say about the voltage across those parallel sets of resistors?

4. Apr 1, 2017

yes.

5. Apr 1, 2017

### Staff: Mentor

Okay, then see @berkeman 's post above.

6. Apr 1, 2017

### Kajan thana

This is the part that I do not understand, all this time I thought that current will change in the whole circuit if the resistance changes where ever it is placed in the ciruit. Then i thought the current is proportional to voltage. so I thought the voltage will also change.

Can you please explain to me where I am going wrong? Thanks.

The actual answer is there will be no change.

7. Apr 1, 2017

### Kajan thana

I actually understand the fact that it will be 6V in each parallel set due to Voltage Law. My problem is where do we use the proportionality rule: If voltage increase the current also increase?

8. Apr 1, 2017

### Staff: Mentor

No. The 12V source is across both the pairs of resistors. The pair of resistors A-E will have 12V across them, and the pair of resistors B-F will have 12V across them.

If the 12V source is ideal (no internal resistance), then the 12V output does not depend on the current being drawn by the load circuit.

9. Apr 1, 2017

### cnh1995

10. Apr 1, 2017

### Staff: Mentor

You are referring to Ohm's law. The voltage involved then is the change in potential across a resistor due to a current flowing through it: V = IR. If there's no resistance along a path then there will be no potential drop due to current flow. In your circuit, the wires are taken to be ideal (no resistance), so there will never be a potential drop between the battery and the branches connected directly to the battery by those wires. That means in this circuit the potential difference (voltage) across both of the branches is fixed at 12 V. To be clear, the potential differences between A and E, and B and F, are both fixed at 12 V by that battery.

11. Apr 1, 2017

### Kajan thana

Sorry, I meant to say 12V, but this is what I don't understand: If I want to find out the voltage across A-C, then I have to apply V=IR , isn't the current determined from the overall resistance?

12. Apr 1, 2017

### cnh1995

It's the "total current supplied by the battery" that is determined by the "overall" resistance.
You need to revise Ohm's law.

13. Apr 1, 2017

### Staff: Mentor

You can look at the total current for that branch alone (Ohm's law for the entire branch), then use that current and Ohm's law on the individual components of the branch.

14. Apr 1, 2017

### Kajan thana

Okay. So I substituted some numbers and I realised the ratio the current divide into each branch causes the voltage to remain same. Am I right in some ways?

15. Apr 1, 2017

### Staff: Mentor

You'll need to be more specific about which current and which voltage. I can identify three separate currents flowing in the circuit.