Answer:
P = 0.25 W
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 :
[tex]P=\dfrac{V^2}{R}\\\\P=\dfrac{2^2}{16}\\\\=0.25\ W[/tex]
So, 0.25 W power is delivered to the bulb.