Question

Two 75.0 W (120 V) lightbulbs are wired in series, then the combination is connected to a 230 V supply.

How much power is dissipated by each bulb? Answer in W.

Answer 1

Power rating = V^2/R
For each, 75.0 = (120)^2/R
R = 192 ohms

Now, as both connected in series,
Req = 192+192 ohms = 384ohms
I = V/Req = 230/384 A = 0.59 A

Power dissipated by each bulb = I^2xR = (0.59)^2 x 192 watt = 66.84W

Answer 2

Calculate the resistance of the bulb by using the relation between power voltage and resistance as follows:

\(\begin{aligned} P &=\frac{V^{2}}{R} \\ R &=\frac{V^{2}}{(P)} \\ &=\frac{(120 \mathrm{~V})^{2}}{(75.0 \mathrm{~W})} \\ &=192 \Omega \end{aligned}\)

Now, these resistors are connected in series then the equivalent resistance becomes:

\(R_{\text {equ }}=R+R\)

$$ \begin{array}{l} =(192 \Omega)+(192 \Omega) \\ =384 \Omega \end{array} $$

Now, these bubs are connected with \(230 \mathrm{~V}\). by using ohm's law the current through these bulbs is:

\(\begin{aligned} V &=I R_{\text {oqu }} \\ I &=\frac{V}{R_{e q u}} \\ &=\frac{(230 \mathrm{~V})}{(384 \Omega)} \\ &=0.599 \mathrm{~A} \end{aligned}\)

Therefore, the required power through each bulb using above current is, \(P=\frac{V^{2}}{R}\)

$$ \begin{aligned} &=\frac{(I R)^{2}}{R} \\ &=I^{2} R \\ &=(0.599 \mathrm{~A})(192 \Omega) \\ &=68.8 \mathrm{~W} \end{aligned} $$