In a factory an electric bulb of 500w
WebAn electric bulb of 60 W is used for 6 h r s / d a y. Calculate the units of energy consumed in one day by the bulb. Calculate the units of energy consumed in one day by the bulb. Q. 16 bulbs of 40w are used for 6 hours a day along with 1 bulb of 100w for 2 hours. WebStep 1: Finding Resistance of bulb using given Ratings Rating of bulb P=100W, V=220V Power, P= RV 2 ⇒ R= PV 2 = 100220×220=484Ω Step 2: Power consumed by bulb Now the voltage drop across the bulb is V=110V So, power consumed by the bulb will be: P b= RV 2= 484110×110 (Resistance of bulb will remain same) P b=25W The correct option is D.
In a factory an electric bulb of 500w
Did you know?
WebAn electric bulb marked 500 W, 250 V is connected to a 250 V supply. a. Find the electric current through the circuit. b. Calculate the resistance of the bulb. Medium Solution Verified by Toppr Given, The power rating of the electric bulb, P=500 W Voltage, V=250 V i) We know, P=VI I= VP I= 250500 I=2 A ii) V=IR R= IV R= 2250 R=125 Ω WebApr 7, 2024 · In this case, it consumes 500 W when a source of 100 V is applied across the bulb. First find the resistance of the bulb using the given rating of the bulb, then consider …
WebNov 26, 2024 · In Q2, if we go 1 step ahead and try to find the voltage across bulb 2, we will find the voltage is = V = IR = 4 A x 8 Ohm = 32 V. Now as per bulb rating the maximum … WebH_ (1)^ (2)+H_ (1)^ (2)->He_ (2)^ (3)+n_ (0)^ (1)+3.27MeV Find the duration for which for which an electric bulb of 500W can be kept glowing by the fusion of 100gm of deuterium. Expert Answer 1st step All steps Final answer Step 1/2 Given Fusion reaction is, 1 2 H + 1 2 H → 2 3 H e + 0 1 n + 3.27 M e V View the full answer Step 2/2 Final answer
WebApr 7, 2024 · Hint:The rating of the bulb tells you that the bulb consumes a particular amount of power when a particular voltage is applied across it.In this case, it consumes $500W$ when a source of $100V$ is applied across the bulb. First find the resistance of the bulb using the given rating of the bulb, then consider the case of $200V$ supply, find … WebAug 30, 2013 · Electric bulb of 500 W used for 2 h per day for 13 days. Electric energy used by the bulb = 500 × 2 × 13 Wh = 13 kWh. Electric motor of 0.5 HP or 372.85 W used for 7 h per day for 13 days. Electric energy used by the electric motor = 372.85 × 7 × 13 Wh = 33.93 kWh. So, total energy used is = 46.93 kWh. Cost per units is Rs 3
WebAn electric bulb rated for 500 W at 100 V is used in a circuit having a 200 V supply. The resistance R that must be put in series with the bulb, so that the bulb draws 500 W, is A 700 Ω B 40 Ω C 20 Ω D 18 Ω Solution The correct option is B 20 Ω Power consumed by a bulb is P = V I,I = P V ⇒ I = 500 100= 5 A
WebNov 6, 2024 · An electric bulb of 500 W at 100v is used in a circuit Engineering Shivang kohlii Nov 4, 2024 electric circuit Nov 4, 2024 #1 Shivang kohlii 19 1 If the circuit has 200 V supply. The resistance R that must be put in series with bulb so that it draws 500 w is? 2. Relevent equations P= v^2/r I = v/r 3. My attempt at the solution birthday post giftsWebApr 22, 2024 · An electric bulb has a rating of 500 W, 100 V. It is used in a circuit having a 200 V supply. What resistance must be connected in series with the bulb so that it deliver 500 W ? (a) 10 Ω (b) 20 Ω (c) 30 Ω (d) 40 Ω amu aiims neet 1 Answer +1 vote answered Apr 22, 2024 by Bhawna (68.7k points) selected Apr 23, 2024 by Farhat Best answer birthday potluck imagesWebZP 2 Pack Led Light Bulb 500W Equivalent 7500 Lumen 5000K E26/E27 Medium Base 60W led Bulb for Outdoor Indoor Lamp Area Garage Warehouse Workshop Street Backyard … birthday powerpointbirthday powerpoint slides freeWeb1 day ago · Find many great new & used options and get the best deals for UFO LED High Bay Light Factory Warehouse Gym Shop Lamp 200W 300W 500W Floodlight at the best online prices at eBay! Free delivery for many products! ... UFO LED High Bay Lights 100W 200W 300W 500W Industrial Factory Workshop Light. £14.56. Free Postage. UFO LED … birthday pound cakeWebQuestion: H_(1)^(2)+H_(1)^(2)->He_(2)^(3)+n_(0)^(1)+3.27MeV Find the duration for which for which an electric bulb of 500W can be kept glowing by the fusion of 100gm of … birthday powerpoint template freeWebJun 4, 2014 · CBSE Class 10 Answered A 500W electric iron used in a house for 2 hours per day. Calculate the cost to use it for 60 days at Rs 3.20 per kWh. Asked by Topperlearning … birthday ppt.pptx