Question

An ammeter with resistance 1.42 Ω is connected momentarily across a battery, and the meter reads 9.78 A. When the measurement is repeated with a 2.11@Ω meter, the reading is 7.46 A. Find (a) the battery voltage and (b) its internal resistance.

Answer 1

Let internal resistance of battery = r

voltage of battery = V

then, in first ammeter, equivalent resistance = Req = r + Ra

here, Ra = resistance of ammeter = 1.42 ohm

So, by ohm's law,

V = I*Req

given, I = 9.78 A

then, V = 9.78*(1.42 + r) eq(1)

in second ammeter, equivalent resistance = Req = r + Ra

here, Ra = resistance of ammeter = 2.11 ohm

So, by ohm's law,

V = I*Req

given, I = 7.46 A

then, V = 7.46*(2.11 + r) eq(2)

by equating both equation,

9.78*(1.42 + r) = 7.46*(2.11 + r)

9.78*1.42 - 7.46*2.11 = (7.46 - 9.78)*r

r = (9.78*1.42 - 7.46*2.11)/(7.46 - 9.78)

r = 0.80 ohm (Part- B)

then, from eq(2),

V = 7.46*(2.11 + 0.8)

V = 21.71 V (Part- A)

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