Answer to Question #124914 in Atomic and Nuclear Physics for Stefanie Hernandez-Mendez

Question

Consider two electric bulbs. Bulb #1 has a resistance of twice as large as bulb #2.
Part (a) These two bulbs are included into a circuit in such a way that the voltage drops across them is equal. What bulb will produce the greater illumination?
The both will produce the same illumination.
It is impossible to answer without actual values of resistances.
Bulb #1
Bulb #2

Now these two bulbs are connected in a circuit so that current flowing through them are the same. Which bulb will produce the greater illumination now?
The both will produce the same illumination.
It is impossible to answer without actual values of resistances.
Bulb #2
Bulb #1

Answer 1

(a) To have equal voltage drops across both bulbs, we must connect them in parallel. The illumination is determined by the power that the bulbs give out:

$$P=\frac{V^2}{R}.$$

For our bulbs, we have

$$P_1=\frac{V_1^2}{R_1}=\frac{V^2}{2R}.\\\space\\ P_2=\frac{V_2^2}{R_2}=\frac{V^2}{R}=2P_1.$$

The second bulb is two times brighter than the first one.

(b) The same current flows when the bulbs are connected in series. The power is current squared times resistance:

$$P=I^2R,\\ P_1=I_1^2R_1=I^2\cdot2R=2I^2R,\\ P_2=I_2^2R_2^2=I^2R.$$

The first bulb will be brighter.