Answer / manish

POWER P = V * I = V * V/R = 100 * 100/(500*1000)
= 10000/500000 = 1/50 = 0.02 WATTS
GIVEN ENERGY = 1 KWH = 1000 WATT HOUR

TIME FOR WHICH BULB WILL GLOW
= (1000 WATT HOUR) / (0.02 WATT)
= 1000/0.02 HOURS
= 50000 HOURS

Answer / chegu

first have to find the I=V/R.V=100V AND R=500k
I=100/500,I=0.5 then find the power consumption
p=v*i,p=500w 1kwh=1000wh ,the power consumption of the bulb
is 500w so it will glow for 2hours according to my
calcution

Answer / k.prakashchandra

500 hours

How to work E.L.C.B and how to connect in circuit.

0 Answers  

---

what is the difference between effective earthing and resistive earthing?

3 Answers   REL,

---

Why we use series reactor in power system?

9 Answers   Cadbury, HEG, Uttam,

---

what is the difference between LLL and LLLG fault ?

1 Answers  

---

Whta is earthing?What is the optimum value of Earth Pit?

2 Answers   Bharti,

---

A heater capacity is 3.0 KW on 415V . if supply voltage is 380Volt . what is heater capacity on 380volt . and how calculate and give formulas . Prakash veer singh Roto power Engg.

5 Answers  

---

why a bulb glow when connected with two end phase and neutral respectively.

1 Answers   ACME,

---

What is meant by cold loop test in instrument

0 Answers   Transnet,

---

Why celling fan rotates in anticlockwise direction and table fan in clockwise direction ? Give technical reason.

5 Answers  

---

what is H.T?specify the range?

6 Answers  

---

what is wave traps and why it is been used in switchyards?

1 Answers   NHPC,

---

when running 2 nos 450kva dg sets in parallel operation ,in one Dg set power factor and Kvar going to lagging and ampere suddenly going high,please give me suggestion to correct the same.

2 Answers  

---