Power consumed by a series combination of appliance:

AS shown in fig. consider a series combination on three bulbs of power P₁, P₂, and P₃; which are operated in the same voltage.

The resistances of the three bulbs will be

As the bulb are connected in series, so their equivalent resistance is

R = R₁ + R₂ + R₃

If P is the effective power of the combination, then

Clearly, when N bulbs of same power P are connected in series, then

As the bulbs are connected in series, the current I through each bulb will be sameAs the bulbs are connected in series, the current I through each bulb will be same

The brightness of the three bulbs will be

P’₁ = I²R₁,             P’₂ = I²R₂,             P’₃ = I²R₃

The bulb of lowest wattage will have maximum resistance and it will glow with maximum brightness. When the current in the circuit exceeds the safety limit, the bulb of lowest wattage will be fused first.

Power consumed by a parallel combination of appliance:

As shown in fig. consider a parallel combination of three bulb of power P1, P2 and P3,: which are operated at the same voltage.

The resistances of the three bulbs will be

Clearly, when N bulbs of same power P are connected in parallel, then

The brightness of the three bulb will be

As the resistance of the highest wattage bulb is minimum, it will glow with maximum brightness. If the current exceeds the safety limit, the bulb with maximum wattage will be fused first. For this reason, the appliance in houses are connected in parallel.

 

As the bulb are connected in parallel, so their equivalent resistance is

Multiplying both sides by V², we get