In the circuit above the 6 Volt battery is shown as a perfect voltage source in

Question

In the circuit above the 6 Volt battery is shown as a perfect voltage source in series with its internal resistance, Rb (both enclosed by the box in dotted lines to represent the complete battery). Observe t the digital multimeter, battery, and the 1 KOhm resistor are all in parallel when the switch is closed. Use Kirchoffs law to derive an expression for the internal resistance of the battery. Your result should be in terms of the 1 KOhm resistance, V0, the open circuit voltage of the battery (when no current flows through battery), and Vc, the closed circuit voltage of the battery (when the switch is closed and a current flows through Rb).

Explanation / Answer

if the switch is open then reading of the multimeter is the emf of the battery =V0 AND WHEN THE SWITCH IS CLOSED the reading of the voltmeter is V0-ir where r= INTERNALRESISTANCE OF THE BATTERY WHERE i=current through the circuit =V0/r+1000 hence reading=V0*1000/r+1000=Vc hence the internal resistance=1000V0/Vc-1000

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