Question

Two identical bulbs of power P are connected in parallel. The third identical bulb of power P is connected in series to this combination. What is the power dissipated in this circuit in terms of P

Answer 1

The power dissipated in this circuit is 3 P.

Step-by-step explanation:

The formula for determining the power in circuit is P = VI, where V is the voltage and I is the current.

So power is directly proportional to the product of voltage and current.

If two bulbs are connected in parallel, then the voltage flow in both the bulbs will be constant. So the sum of varying current in the circuit will be

`I_(p)= I_(1)+ I_(2)`

As here
`I_(p)=(P_(p) )/(V_(p) )` ,
`I_(1)=(P_(1) )/(V_(1) )` and
`I_(2)=(P_(2) )/(V_(2) )`

Since , the voltage is similar in parallel connection, Vp=V1=V2=V

So,
`(P_(p) )/(V)=(P_(1) )/(V)+(P_(2) )/(V)`

Then
`P_(p) = P_(1)+ P_(2)`, as both the bulbs are identical bulbs then
`P_(p) = 2 P`

Then , along with these two another bulb is connected in series. So in series , the current will be constant, then

`V_(s) = V_(1) + V_(2)`

And
`V_(s)=(P_(s) )/(I)`,
`V_(1)=(P_(p) )/(I)` and
`V_(2)=(P_(3) )/(I)`

`(P_(s) )/(I)=(P_(p) )/(I)+(P_(3) )/(I)`

Since the bulb is identical so P3 =P

`P_(s) = P_(p)+ P_(3)=2P+P = 3P`

So the power dissipated in this circuit is 3 P.