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

## Homework Statement

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

V=IR[/B]

## The Attempt at a Solution

:[/B]
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.

#### Attachments

• Screen Shot 2017-04-01 at 18.59.47.png
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gneill
Mentor
When you specify "A-C", are you asking about the potential difference across the upper 20 kΩ resistor?

• Kajan thana
berkeman
Mentor

## Homework Statement

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

V=IR[/B]

## The Attempt at a Solution

:[/B]
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.
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?

• Kajan thana and cnh1995
When you specify "A-C", are you asking about the potential difference across the upper 20 kΩ resistor?
yes.

gneill
Mentor
yes.
Okay, then see @berkeman 's post above.

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?
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.

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?
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?

berkeman
Mentor
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?
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.

• Kajan thana
cnh1995
Homework Helper
Gold Member
If the voltage across a resistor increases the current through the resistor also increases

• Kajan thana
gneill
Mentor
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?
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.

• Kajan thana
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.
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?

cnh1995
Homework Helper
Gold Member
isn't the current determined from the overall resistance?
It's the "total current supplied by the battery" that is determined by the "overall" resistance.
You need to revise Ohm's law.

• Kajan thana
gneill
Mentor
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?
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.

• Kajan thana
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.
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?

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