Question

a typical cmall flashlight contains two batteries each having na emf of 2.0 v connected in series with a bulb havin ga resistance of 16 ohms if the internal resistance of the batteries is negligible what power is delivered to the ublb

Answer 1

P = 0.25 W

Step-by-step explanation:

Given that,

The emf of the battry, E = 2 V

The resistance of a bulb, R = 16 ohms

We need to find the power delivered to the bulb. We know that, the formula for the power delivered is given by :

`P=(V^2)/(R)\\\\P=(2^2)/(16)\\\\=0.25\ W`

So, 0.25 W power is delivered to the bulb.