Question

A battery of e.m.f. 4V and internal resistance 2 ohms is joined to a resistor of 8 ohms. What additional resistance in series with 8 ohms resistor would produce a terminal p.d. of 3.6V?​

Answer 1

Explanation:

w.k.t

I=E/R+r

V=IR=ER/R+r

3.6=4(R'+8) / (R'+8)+2

3.6R' +36=4R'+32

0.4R'=4

R'=10 ohm