Assume that you have a battery of emf e m f and three identical lightbulbs, each having constant resistance r. (a) what is the total power from the battery if the bulbs are connected in series? (use the following as necessary: ℰ for e m f and r.) (b) what is the total power from the battery if the bulbs are connected in parallel? (c) for which connection will the bulbs shine the brightest? parallel series

Explanation:

The power consume by an electric resistance Rt is :

I²*Rt  =  P    (in watts)   R is in  Ω , and V or  E  in volts

V* I  = P is also a valid expression for the power

a) What is the total power from battery, if the bulbs are connected in series

if   R₁  R₂  and R₃  three resistors are connected in series,  the equivalent resistor is the sum of the three resistor. In this particular case resistors are identical then Rt = R + R + R    Rt = 3R

Then Power will be:

P  (w)   =  V* I         and according to ohm´s law   V =  I*R   then  V/R = I

P  (w)   = V*V/Rt     P  (w)   = V² /Rt         P  (w)   = V² /3R  

b) In parallel connection

The equivalent  resistor should be compute according to

1/Rt  = 1/R₁  +  1/R₂  +  1/R₃    again three equal resistor R

1/Rt   =  1/R  +  1/R  *  1/R        1/Rt   =   3/R

Then     Rt  = R/3

Power:     P =  V²/Rt    By subtitution     P = V²/Rt    P = V²/R/3

P  =  3 V²/R

In this case bulbs shine the brightest.

Note: Lighting circuits should be installed in parallel. Suppose the bulbs are rated 120 (V) and the source is 120 (V)   if you have the three bulbs in series the drop voltage for operation of the first bulb will force the second one, an operation with less than its rated voltage and the situation is worse for the third ( it maybe dont light at all)