Question

Two light bulbs are connected in parallel, and then connected to a battery, as shown in Fig. 4.51(a). You observe that bulb 1 is twice as bright as bulb 2. Assuming that the brightness of a bulb is proportional to the power dissipated in the bulb’s resistor, which bulb’s resistor is larger, and by what factor? The bulbs are now connected in series ,as shown in Fig.4.51(b). Which bulb is brighter, and by what factor? How bright is each bulb compared with bulb 1 in part (a)?

(a) (b) Figure 4.51.

Answer 1

a) let battery voltage = V

power dissipation in bulb 1 is P1 = V^2 /R1

and power dissipated in bulb 2 is P2 = V^2/R2

=> P1/P2 = R2/R1

=> 2 = R2/R1

=> R2 = 2*R1 , bulb 2 resistor is 2 times the bulb 1 resistor.

b) in series power dissipation = I^2*R

so P1 = I^2*R1 , P2 = I^2*R2

=> P1/P2 = R1/R2 = 1/2

=>   P2 = 2*P1

=> bulb 2 is brighter by factor of 2.