When the current through an ohmic conductor is 2A, the potential difference across it is 6V. What is the potential difference across the same ohmic conductor when the current is increased to 3A?

To answer this, you need to pick up on the clues in the question. "Ohmic conductor" is explicitly mentioned twice, so this is likely to be important. You should know that for ohmic conductors, p.d. is proportional to current, which means p.d./current=constant (called the constant of proportionality). You can calculate the constant from the first part of the question, 6/2=3.Now you know the constant, rearrange the equation to make what you want to find out, the p.d., the subject:p.d.=constantcurrentNow, you can use this equation and put in the numbers you know: p.d.=33=9V (don't forget the units!)If you'd prefer, you can also solve this using a graph. Draw a straight-line graph that cuts through zero-zero, with p.d. on the y-axis and current on the x-axis. The constant of proportionality is the gradient of the line, and can be found by finding difference in two y values (6-0) and dividing by difference in two x values (2-0). You can then use the equation above to find the p.d. in the second scenario, and you should get the same answer, 9V.

Answered by Eleanor Grace G. Physics tutor

3127 Views

See similar Physics GCSE tutors

Related Physics GCSE answers

All answers ▸

What is an example of a natural satellite?


What is the evidence for the Big Bang theory?


A person swims from a depth of 0.50 m to a depth of 1.70 m below the surface of the sea. Density of the sea water = 1030 kg/m^3. Gravitational field strength = 9.8 N/kg. Calculate the increase in pressure on the swimmer. Give the unit.


A student of mass m=50kg runs an experiment. He throws a ball of mass m = 400g from a height h = 20m. What will be the speed of the ball he records just before it touches the ground?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences