Question:
a battery of e.m.f 3 volt and internal resistance r is conected in series with a resistor of 55 ohms through of an ammeter of resistance 1 ohm .the ammeter reads 50mA .draw the circuit diagram and calculate the value of r
Answer:
The battery is connected with resistance 50 Ω and ammeter in series.
Given, internal resistance of battery= r Ω
Resistance of ammeter = 1 Ω
So,total resistance in circuit, R= (r+55+1)Ω
= (r+56) Ω
Current,I= 50 mA = 50×10-3 A
EMF, V= 3 volt
From, V=IR
3 =50×10-3 ×(r+56)
r = 4 Ω
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