Question

2.4 You decide to explore the great outdoors and go camping, but you find you forgot fuel for your lantern. It is really dark, so you decide to feel around in the bag you hastily packed to make a light. You find a 9.0V battery, bulbs, and wires, but the bulbs available are rated for 3.0V and are rated 2.5 Watts at that voltage. The bulb will burn out very quickly if it experiences more than a 3.0 V potential drop across it. You want to calculate the resistor you need to add to the circuit so you won't burn out the bulb. You need to calculate this in advance because you want to waste as few matches in your pocket as possible, and you will need a match to look through the package of resistors you packed (why did you pack those). What value resistor do you need? (10 points) How much power will the resistor dissipate? (5 points) R HE 9.0 V

Answer 1

Given that the power of the bulb P = 2.5 watt and maximum voltage it can withstand V = 3.0 V

so current through the bulb

power I Voltag e
$= \frac{2.5 watt}{3.0 volt}$

$= \frac{5}{6} ampere$

since applied voltage V = 9.0 V,

sum of voltage dropped across the resistor and bulb is equal to the apllied voltage

IR= 6.0

$R =\frac{6.0}{I}$

$R = \frac{6.0}{\frac{5}{6}}$

$=\frac{36}{5}$

is the resistance of the resistor needed

Power dissipated is P = I²R

= (5/6)² (36/5)

= 5 watt