Question

A battery rated at 7.2 V and 10000 J is connected across a light bulb. Assume that the internal resistance of the battery is negligible, and that the resistance of the bulb's filament is 100 Ω. Draw the associated circuit, and label all branch variables, following the Associated Variable Convention. What is the power into the bulb? What is the power into de battery? How long will the battery last in the circuit?

Answer 1

Answer:

b) Power to bulb = 0.5184 watt

Power into the battery = - 0.5184 watt

c) T = 19290.123 sec

Explanation:

Given data:

Voltage = 7.2 V

energy = 10,000 J

Resistance = 100 ohm

a) associated circuit is shown in attachment

b)$Power = \frac{v^2}{R} = \frac{7.2^2}{100}$

Power = 0.5184 watt

Power into the battery = - 0.5184 watt

c)

$Power = \frac{E}{t}$

$T = \frac{E}{P}$

$T = \frac{10000}{0.5184}$

T = 19290.123 sec