Question

answer for 9th question...​

Answer 1

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 2

$\tt{\huge{\pink{Hello!!! }}}$

Question :

A bulb is rated at 330 V - 110 W.

What is it's resistance?

Three such bulbs run for five hours at a stretch. What is the energy consumed?

Calculate it's cost In rupees if it's rate is 70 Paisa per unit .

Solution :

$\tt{ \purple{ \implies{r \: = \dfrac{{v}^{2}}{p} \: = 990 \: ohm . }}}$

Considering that the 3 bulbs were used at specified Voltage of 330 V for 3 hours, we get :

$\tt{\red{\therefore{E = 3 \times (Energy \: of \: 1 \: bulb)E=3×(Energyof1bulb)}}}$

$\tt{\green{\implies{ E = 5940 \: Kj }}}$