Question

1.An electric geyser rated 1.1Kw, 220V is used 5hr a day. Calculate:
a) The current drawn
b) its resistance
c) energy consumed by it in 10days
d) minimum rating of fuse that can be safely used.​

Answer 1

Answer:

Power=1.1k

V=220V

Time=5hr

Explanation:

PART-1: :

Power = Current x Potential Difference

P = IV

1100kw = I x 220

I = 5 amperes

PART-2 :

P = I^2 x R

P/I^2 = R

1100/25 = R

R = 44 ohms

PART-3 :

Power consumed in 5 hour = 1100kw

Power consumed in 1 hour = 1100/5 = 220kw

No. of hour in one day = 24hrs

No. of hours in 10 days = 24 × 10 = 240 hr

therefore, Energy = P×T

E = 220 × 240

E= 52800 Kwh

Part-4 :

Power: 1.1 kW

Voltage: 220 V

current = power/voltage = 1, 100 W / 220 V = 5 Amp

The current that passes through the circuit having the geyser is 5 Amp. Then the rating of the fuse must be more than 5 A.