Question

A cell of e.m.f. 1.8 V and internal resistance 2 ohm is connected in series with an ammeter of resistance 0.7 ohm and a resistor of 4.5 ohm. (A) What would be the reading of the ammeter?
(B) What is the potential difference across the terminals of the cell?

Answer 1

Dear Student
Ammeter resistance is added with the circuit resistance and it decrease the e.m.f.
In the given question ammeter's equivalent resistance = 0.7 ohm resistor = 4.5 ohm total resistance= 5.2 ohm
And total circuit resistance = 2+5.2 = 7.2 ohm
now current (I) = V/R = 1.8/7.2 = .25 ampere = 250 milliampere
The potential difference is = .25*7.2 = 1.8 v


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