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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