Question

Each of three bulb having same resistance can dissipate maximum power P0 The maximum power the circuit of three bulbs can dissipate is​

Answer 1

Answer:

Each board has three 40-watt bulbs, connected as shown by the resistor circuit painted on it. The board on the left has the bulbs arranged, of course, in parallel, and the board on the right has them in series. Since power, P, equals iV, P/V = i, so at 120 V, a 40-watt bulb draws 1/3 A. (The units in iV are (C/s)(N-m/C), or J/s, which are watts.) For a given resistance, V = iR, so the bulb’s resistance (when it has 120 volts across it) is 120/(1/3), or 360 ohms. (We also know by the two equations above that P = i2R, which gives R as 40/(1/9), or 360 ohms.)

When the bulbs are connected in parallel, each bulb has 120 V across it, each draws 1/3 A, and each dissipates 40 watts. In this circuit, all bulbs glow at their full brightness. The total power dissipated in the circuit is three times 40, or 120 watts (or 3(1/3) A × 120 V = 120 W).

Explanation:

Correct answerer is

27W

equivalent resistance from A→B

(R∥R)+R

⇒

R

1

+

R

1

1

+R=

R

2

1

+R=

2

R

+R

=

2

3R

∴ equivalent resistance =

2

3R

Power across the single resistor =18 W

×18=27 W ∴ maximum

power consumed by circuit =27 W