Question

A cell of EMF 10V and internal resistance 2 ohms is connected across a resistance 8 ohms. Find the potential difference across battery.

Answer 1

Answer:

potential difference across battery is 8 volt

Explanation:

we can see in question diagram the resistance R= 8 ohm and internal resistance r= 2 ohm are in series connection.

so the net resistance of circuit is

Rnet = R + r

Rnet = 8+2 = 10 ohm

now the current flowing in given circuit can be find

by ohms law E=IR

I= E / R      emf of battery E = 10 volt given in question

I = 10 / 10

I = 1 Amp

now we can find terminal difference across battery

V = E - Ir

V = 10 - 1×2

V = 10 - 2

V = 8 volt .