Question

an electric bulb is rated at 200V ,100W .what is the resistance?if 5 such bulbs burn for 4 hours .what is the electrical energy consumed?calculate the cost if ,the rate is 50 paise per unit.

Answer 1

p=v²/R

100=200×200/R

R=400ohm

E=Pt

=100×4×5

=2000wh

cost = 2000×50/100

=1000

Answer 2

Given :

Power (P) = 100 W

Potential difference (V) = 200 V

To Find :

(i) Resistance of the bulb

(ii) Electrical energy consumed

(iii) Cost if the rate is 50 paise per unit.

Solution :

(i) Resistance of the bulb (R) = $\frac{V^2}{P}$

                                                =  $\frac{200 * 200}{100}$

                                                =  400 ohm

∴ The resistance of electric bulb is 400 ohm.

(ii) Energy consumed by each bulb = Power(P) × Time (T)

                                                            = 100 × 4

                                                            =  400 WH

                                                            =  0.4 KWH

Energy consumed by 5 bulbs = 5 × 0.4 = 2 KWH

∴ The electrical energy consumed by 5 bulbs is 2 KWH

(iii) Cost of operation at 50 paise per unit = 2 × 50

                                                                     =  100 paise

                                                                     =  Rs. 1

Cost of operation for 30 days = 1 × 30

                                                  = Rs. 30

Therefore, the cost of operation of 5 bulbs is Rs. 1 per day and Rs. 30 for 30 days.