A flashlight uses two 3V batteries in series to provide a cu

A flashlight uses two 3-V batteries in series to provide a current of 0. 7 A in the filament, (a) Find the potential difference across the flashlight bulb, Calculate the resistance of the filament.

Solution

Now as the figure is Not given, Assuming the Circuit having all the elements to be connected in Series,we have a flashlight and two 3 Volt Batteries, having current 0.7A.

a)Now as the batteries are connected in Series, and as both are of 3 Volts Each so we will get a Potential difference of 6 Volt Across the flashlight.

b)The Current in the wire is 0.7A and the Voltage is 6 Volts, taking the resistance of the filament to be \'R\' so applying Kirchoffs laws, we get an equation 0.7R=6.So R=6/0.7=8.571 Ohm.

So R=8.571 Ohm is the Resistance of the Filament of the flashlight.

Thank You