Two bulbs rated (P1, V) and (P2, V) are connected (i) in series and (ii) in parallel across supply V. Determine power in the two combinations P1 and P2.

Electric power can be defined as the rate at which work is done or energy is converted into an electric circuit. In other words, it is the measurement energy used in a span of time.

Step 1:Formula

Electric Power formula is given by:

P = VI

Here (in the circuit),

  • P = Power
  • V = Potential Difference
  • I = Electric Current

Power is often also denoted as,

⇒ P = I2R
⇒ \(P = {{V^2} \over R}\)

\(\therefore\) Expressions above are determined using Ohm’s Law, wherein, Voltage, current, and resistance are interrelated by the relation:

V = IR

Here (in the circuit),

  • R = Resistance
  • V = Potential Difference 
  • I = Electric Current

Step 2: Solution

(i) Two bulbs of (P1, V) and (P2, V) that are connected in series:

The flow of current in both bulbs is the same. Assuming R1 and R2 are both resistances of the bulbs, then the effective resistance circuit:

⇒ R = R1 + R2

Now, after using the relation, we can obtain:

P=V2/R

And, R = V2/P

Therefore, V2/P = V12/ P1 + V22/P2

As is given, V1 = V2 = V.

Therefore,

⇒ 1/P = 1/P1 + 1/P2

Hence, effective power of two bulbs connected in series is 1/ P = 1 / P1 + 1 / P2

(ii) Two bulbs of (P1, V) and (P2, V) when connected in parallel:

1/R = 1/R1 + 1/R2

We know that

P = V2/R

1/R = V2/P

As is given, V1 = V2

Thus, P/V2 = P1/V2 + P2/V2

Hence, the effective power of two bulbs connected in parallel is P = P1 + P2.


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