Figure shows an arrangement to measure the emf ϵ and internal resistance r of a battery. The voltmeter has a very high resistance and the ammeter also has some resistance. The voltmeter reads 1.52 V when the switch S is open. When the switch is closed the voltmeter reading drops to 1.45 V and the ammeter reads 1.0 A. Find the emf and the internal resistance of the battery.
1.52V, 0.07Ω
Given,
Voltage reading, V1 with switch is open=1.52V
Voltage reading, V2 with switch is closed=1.45V
Current through the ammeter, i= 1A
Formula used
Kirchhoff’s loop rule says that the algebraic sum of the voltage in a loop is always zero. Or
For a cell with an internal resistance r and emf ϵ, the voltage drop, V across it, if a current ‘i’ is passed through the circuit can be written using the loop rule as,
Solution:
a) When switch is open, the current would circulate through loop ABCA but not through the loop ACDEA. Since the internal resistance is very small compared to the resistance of Voltmeter, the voltage drop occurs completely across the voltmeter. This voltage drop will be measured in the meter and it will be almost equal to the emf of the cell.
Hence the e.m.f of cell= volt meter reading.
Or,
b) When the switch is closed, a current ‘i’ will flow through the loop ACDEA. Now the volt meter will show the potential drop across the cell and the internal resistance combined. So, using eqn.1 , we can find the internal resistance,
Where V=Volt meter reading= 1.45V, i= Ammeter reading=1A, and ϵ= 1.52V as we
By re-arranging, we can find the expression for internal resistance as,
By substituting the given values,
Hence the internal resistance of the cell is 0.07Ω