A battery of emf 2.5 V and internal resistance r is connected is series with a resistor of 45 ohm through an ammeter of resistance 1 ohm. The ammeter reads a current of 50 mA. Draw the circuit diagram and calculate the value of r.

Explanation

Circuit diagram for the given data is shown below.

Given, `E = 2.5 V, R = 4 5 Omega`,
`r_(A) = 1A, I = 50 mA`
r = ?
`E = I (R + r_(A) + r)`
`2.5 = 50 xx 10^(-3) (45 + 1 + r)`
`46 + r = (2.5)/(50 xx 10^(-3)) = (2.5 xx 10^(3))/(50) = 50`
`:. r = 50 - 46 = 4 Omega`