Answer:
The total power dissipated in this circuit is 3P/2
Explanation:
Given that the power of each bulb is P
Let R represents their resistance
From V = IR --- make I the subject of formula
I = V/R
and P = VI
where V = Voltage, I = Current, R = Resistance and P = Power
So, P = V²/R ---- (1)
From the question, we understand that two bulbs are connected serially.
This means that they operate at the same voltage..
So, the total resistance for bulb 1 and 2 is R + R = 2R
Using I = V/R
For bulb 1 and 2
I = V/2R.
So, their power, P is calculated as
P1 = VI
P1 = V * V/2R
P1 = V²/2R
From (1) above, we noted that P = V²/R
So,
P1 = ½V²/R
P1 = ½P.
The power in bulb 1 and 2 is ½P.
The third build is connected in parallel to the first two bulbs.
So the total power is calculated as
½P + P
= P(½+1)
= P(3/2)
=3P/2
