A battery of emf 10V and internal resistance 3Ω is connected to a resistor. The current in the circuit is 0.5A

Emf of the battery, E=10V

Internal resistance of the battery, r=3Ω

Current in the circuit, I=0.5A

Resistance of the resistor = R

The relation for current using Ohm’s law is,

I = E/(R+r)

⇒ R+r = E/I

By substituting E = 10V and I = 0.5A

We get,

⇒ R+r = 10/0.5

⇒ R+3 = 20

⇒ R = 20 – 3

⇒ R = 17Ω

Terminal voltage of the resistor = V

According to Ohm’s law,

V = IR V = 0.5 x 17

We get,

V = 8.5V

Hence, the resistance of the resistor is 17Ω and the terminal voltage is 8.5V

Was this answer helpful?


0 (0)

Upvote (0)

Choose An Option That Best Describes Your Problem

Thank you. Your Feedback will Help us Serve you better.

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *




Free Class