Two bulbs are rated (P1, V) and (P2, V). If they are connected (i) in series and (ii) in parallel across a supply V Find the power dissipated in the two combinations in terms of P1 and P2.

 Electric power is the rate at which work is done or energy is transformed into an electrical circuit. Simply put, it is a measure of how much energy is used in a span of time.

Formula

The formula for electric power is given by

P = VI
where,

P is the power

V is the potential difference in the circuit

I is the electric current

Power can also be expressed as

P = I2R
P = V2/ R
The above two expressions are got by using Ohm’s law, Where, Voltage, current, and resistance are related by the following relation

V = IR
Where,
R is the resistance in the circuit.
V is the potential difference in the circuit
I is the electric current

Solution

(i) When two bulbs rated (P1, V) and (P2, V) are connected in series:

The current is the same via each bulb. If R1 and R2 are resistances of two bulbs, then the effective resistance of the circuit is

R = R1 + R2

Using the relation we get

P=V2 / R and 

R = V2 / P

Hence 

 V2 / P = V12 / P1 + V22 / P2

Given: V1 = V2 = V.

Therefore,

1/ P = 1 / P1 + 1 / P2

Thus, the effective power when two bulbs are connected in series is

1/ P = 1 / P1 + 1 / P2

(ii) When two bulbs rated (P1, V) and (P2,V) are connected in parallel

1/R = 1/R1 + 1/R2

We know that

P = V2/ R

1/R = V2/ P

Given V1 = V2

Hence

P/V2 = P1/V2 + P2/V2

Thus, the effective power when two bulbs are connected in parallel is P = P1 + P2.