Question

Which has greater resistance: a 75 watt bulb or a 100 watt bulb?

Suppose you connect a 75 watt bulb across only 50 volts: will you still get 75 watts of power?

Suppose you connect a 100 watt bulb and a 75 watt bulb in series and then connect the combination to a regular 110 volt line. Which bulb (if either) will burn brighter?

(note that we are talking about the older incandescent bulbs here... the newer types are quite different.)

Answer 1

Power = V^2/R
If the Voltage source is constant, Then 100 Watt Bulb will have Smaller resistance while 75 Watt Bulb will have Greater resistance.

No, power rating of bulb depends on the Voltage supplied, if we lower the voltage it won't glow at max Power.
So there must also be a voltage rating along with the wattage to be meaning full.

Same current will flow through both the bulb's but as we discussed above 75 Watt Bulb will have Greater resistance, So The 75 W bulb will burn brighter than the 100 W bulb.

Answer 2

The power across \(75 \mathrm{~W}\) bulb is,

$$ \begin{array}{l} P_{1}=\frac{V^{2}}{R_{1}} \\ R_{1}=\frac{V^{2}}{P_{1}} \\ =\frac{V^{2}}{75 \mathrm{~W}} \\ P_{2}=\frac{V^{2}}{R_{2}} \\ R_{2}=\frac{V^{2}}{P_{2}} \\ =\frac{V^{2}}{100 \mathrm{~W}} \end{array} $$

Since \(_{\text {2n }} R_{1}>R_{2}\), the resistance of \(75 \mathrm{~W}\) bulb is more than the resistance of \(100 \mathrm{~W}\) bulb.

If we connect a \(75 \mathrm{~W}\) bulb across \(50 \mathrm{~V}\),

$$ \begin{aligned} R_{1} &=\frac{V^{2}}{75 \mathrm{~W}} \\ &=\frac{V^{2}}{75 \mathrm{~W}} \\ I &=\frac{(50 \mathrm{~V})}{\frac{V^{2}}{75 \mathrm{~W}}} \\ &=\frac{(50 \mathrm{~V})(75 \mathrm{~W})}{V^{2}} \end{aligned} $$

Power,

$$ \begin{aligned} P &=V_{\text {supply }} I \\ &=\frac{(50 \mathrm{~V})^{2}(75 \mathrm{~W})}{V^{2}} \end{aligned} $$

The power rating depends on the voltage supplied. More the voltage more the rating.

Since, the resistance of the \(75 \mathrm{~W}\) lamp is more than the resistance of the \(100 \mathrm{~W}\) lamp, and same current will be flowing through both the resistors connected in series, then according to the formula \(P=I^{2} R,\) the \(75 \mathrm{~W}\) lamp would bum brighter than the \(100 \mathrm{~W}\) lamp.