Answer for 9th question...

1.	Answer for 9th question...
Answer» Answer: Given: Power rating of a bulb is 330V - 110W Also the 3 bulbs are used for 5 hrs at a stretch. To find: Resistance of bulb , and energy consumed. Calculation: Power rating of any electronic DEVICE provides the ideal power GENERATED when run on a specified Voltage. In this case , when the bulb is used at 330V , we get power GENERATION of 110W. Let power be P , voltage be V and resistance be R . $P = \dfrac{ {V}^{2} }{R}$ $= > 110 = \dfrac{ {(330)}^{2} }{R}$ $= > R = \dfrac{ {(330)}^{2} }{110}$ $= > R = 990 \: ohm$ Considering that the 3 bulbs were used at specified Voltage of 330 V for 3 HOURS, we get : $E = 3 \times (Energy \: of \: 1 \: bulb)$ $= > E = 3 \times \{110 \times(5 \times 60 \times 60) \}$ $= > E = 5940 \: kJ$

Answer»

Answer:

Given:

Power rating of a bulb is 330V - 110W

Also the 3 bulbs are used for 5 hrs at a stretch.

To find:

Resistance of bulb , and energy consumed.

Calculation:

Power rating of any electronic DEVICE provides the ideal power GENERATED when run on a specified Voltage. In this case , when the bulb is used at 330V , we get power GENERATION of 110W.

Let power be P , voltage be V and resistance be R .

$P = \dfrac{ {V}^{2} }{R}$