Question

Usually wire resistance can be neglected. But we see the effect of it sometimes when a light dims as another high-current device is turned on. In the fridge, the bulb will put out a little less power when the switch to the compressor motor is turned on.
a. The voltage source (wiggly circled curve) is 120 V (treat it like a battery) and the wire resistance is 0.5 ohms. The bulb is rated as "75W", which means its power is 75W when connected to 120 V. What is the resistance of the bulb? (ohms)
b. When the switch to the motor is open (preventing that branch of the circuit from receiving current, so you can pretend it is not even there), find the power consumed by the bulb, which should be a little lower than 75W because of the wire resistance. (W)

Answer 1

Please find the figure attached below:

Answer:

a.Resistance of bulb=$R_{2}$=192 ohm

b.power consumed by bulb after motor disconnection=74.609 W

Explanation:

a.Resistance of bulb=$R_{2}$=?

As we know that $P=\frac{P^{2} }{R}$

putting R as resistance of bulb i.e. $R=R_{2}$

$P=\frac{V^{2}}{R_{2} }\\R_{2}=\frac{V^{2}}{P} \\ R_{2}=\frac{(120)^{2} }{75}\\ R_{2}=192ohm$

b.power consumed by bulb after motor disconnection?

from the figure we see that $R_{1}\ and\ R_{2}$ are in series so

$R_{eq}=R_{1} +R_{2} \\R_{eq}=192+0.5\\R_{eq}=192.5ohm$

current through their resistance is

$I_{eq}=\frac{V}{R_{eq} }\\ I_{eq}=\frac{120}{192.5 } \\I_{eq}=0.6233A$

Power consumed by bulb is

$P_{b} =I^{2}R\\\\ P_{b} =(0.6233)^{2}(192)\\\\P_{b} =74.609W$