Question

#25 points ! please answer very very fast ! toooooo urgent !

Answer 1

1) In the first case,
Units of electrical energy consumed = 100W*4bulbs*5hours*30days = 60000 WH = 60 KWH = 60 units
After replacing , Energy consumption = 8W *4 bulbs*5hours*30 days = 4800 WH = 4.8 KWH = 4.8units
Units of energy saved = 60-4.8 = 55.2.

Answer 2

1. bulb: 
P= 100W = 0.1kW
​t = 5* 30 = 150 hr
E = P*t
   = 0.1 * 150 = 15
Power for 4 bulbs = 15* 4 = 60
LED:
P = 8W = 0.008
t = 5 * 30 = 150hrs
​E = 0.008 * 150
   = 1.2
Power for 4 LED = 4 * 1.2 = 4.8
Power saved per month = 60 - 4.8 = 55.2
so option (a) is correct

2. according to the given circuit:
resistance are connected in parallel.
So effective resistance is 1/R=1/R1+1/R2+1/R3
∴1/R=1/2+1/4+1/6
⇒11/12
∴r=12/11 ohms.
Voltmeter reading:
V=IR⇒11x12/11= 12 volts[∵i=11 amp]

Ammeter reading:
V=IR
I=V/R
=12/4
3 amp [∵R=3 ohms]
So voltmeter reading is 12volts and ammeter reading is 3amp