Q. 16

# Figure shows an a

Answer :

1.52V, 0.07Ω

Given,

Voltage reading, V_{1} with switch is open=1.52V

Voltage reading, V_{2} 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

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