You are Here: Home >< Physics

# Help with potential divider question ? watch

1. I'm only stuck on one part at the end but I will write out the full question:

A potential divider consists of an 8.0 resistor in series with a 4.0 resistor and a 6.0V battery of negligible internal resistance. Calculate
i) the current (done)
ii) the pd across each resistor (done)

b)
In the circuit above the 4 resistor is replaced by a thermistor with a resistance of 8 at 20C and a resistance of 4 at 100C. Calculate the pd across the fixed resistor at
i) 20 C (done) (answer 3V)
ii) 100C

I did for B ii) 6.0x4.0 / 12 and I get the answer of 2V yet the textbook says the answer is 4V but i've got the questions right so far, is this a textbook error or am I not seeing something.

Thanks
2. (Original post by Cinna21)
I'm only stuck on one part at the end but I will write out the full question:

A potential divider consists of an 8.0 resistor in series with a 4.0 resistor and a 6.0V battery of negligible internal resistance. Calculate
i) the current (done)
ii) the pd across each resistor (done)

b)
In the circuit above the 4 resistor is replaced by a thermistor with a resistance of 8 at 20C and a resistance of 4 at 100C. Calculate the pd across the fixed resistor at
i) 20 C (done) (answer 3V)
ii) 100C

I did for B ii) 6.0x4.0 / 12 and I get the answer of 2V yet the textbook says the answer is 4V but i've got the questions right so far, is this a textbook error or am I not seeing something.

Thanks
you might be calculating the PD across the other resistor by accident
3. (Original post by Joinedup)
you might be calculating the PD across the other resistor by accident
Are you sure ? did you work out the question and get 4 ? please help I'm confused
4. (Original post by Cinna21)
Are you sure ? did you work out the question and get 4 ? please help I'm confused
at 100 deg C you've got an 8 ohm fixed resistor in series with a 4 ohm thermistor

across both resistors in series you've got the supply voltage (6V)

across the 8 ohm you've got supply voltage x 8/(8+4)
across the 4 ohm you've got supply voltage x 4/(8+4)

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: February 23, 2016
Today on TSR

### Exam Jam 2018

Join thousands of students this half term

### Solo - A Star Wars Story

Poll

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE