Consider two circuits:(i) A: in which N identical bulbs are connected ...
Explanation:
Circuit A:
- In circuit A, N identical bulbs are connected in series across a battery of emf E.
- When bulbs are connected in series, the total resistance of the circuit increases as compared to a single bulb.
- As a result, the current flowing through each bulb decreases.
- The power dissipated in each bulb can be calculated using the formula: P = I^2 * R, where P is power, I is current, and R is resistance.
- Since the current decreases in a series circuit, the power dissipated in each bulb in circuit A (PA) will be less than the power dissipated in a single bulb (P) when it is connected individually across the same battery.
- Therefore, PA < />
Circuit B:
- In circuit B, N bulbs identical to those in circuit A are connected in parallel across a similar battery of emf E.
- When bulbs are connected in parallel, the total resistance of the circuit decreases as compared to a single bulb.
- As a result, the current flowing through each bulb increases.
- The power dissipated in each bulb can be calculated using the formula: P = I^2 * R, where P is power, I is current, and R is resistance.
- Since the current increases in a parallel circuit, the power dissipated in each bulb in circuit B (PB) will be greater than the power dissipated in a single bulb (P) when it is connected individually across the same battery.
- Therefore, PB > P.
Total Power Delivered:
- The total power delivered by the battery in circuit A (PAT) can be calculated by multiplying the power dissipated in each bulb (PA) by the number of bulbs (N).
- Therefore, PAT = N * PA.
- The total power delivered by the battery in circuit B (PBT) can be calculated by multiplying the power dissipated in each bulb (PB) by the number of bulbs (N).
- Therefore, PBT = N * PB.
Conclusion:
- From the above analysis, we can conclude that the power dissipated in each bulb in circuit B (PB) is greater than the power dissipated in each bulb in circuit A (PA).
- Hence, option 'D' is the correct answer: PB = N^2 * PA.
Consider two circuits:(i) A: in which N identical bulbs are connected ...
∴ We get P
E = N
2P
A
To make sure you are not studying endlessly, EduRev has designed Class 10 study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Class 10.