Web25W Medium Solution Verified by Toppr Correct option is B) A 60 Watt 220V bulb means that it will consume 60 Watts when the voltage across it is 220V. In such a case the current flowing through it will be given by I=W/V=60/220 Amperes. And the resistance of the filament will be W/I 2=60×( 60220) 2=806.67 ohms WebSolution 25W = P1 and 200 W = P2 are the ratings on the bulbs. That means those bulbs are supposed to work at that power ratings. Voltage ratings of hte bulbs are: 120Volts. …
Two electric bulbs marked 25W-220V and 100W-220V are …
WebResistance of 100 w bulb= (200)^2/100=400 ohm Total resistance= 1000+666.67+400=2066.67 ohm ( addition because series circuit) Now, we calculate current flowing in the bulbs ( as we know this is series circuit so same current will flow in the circuit) Current (I)= Voltage/total resistance=200/2066.67=0.0967 A WebSolution 25W = P1 and 200 W = P2 are the ratings on the bulbs. That means those bulbs are supposed to work at that power ratings. Voltage ratings of hte bulbs are: 120Volts. R1 and R2 are their resistances. P = V² / R => R1 = V² / P1 = 120² / 25 = 576 Ohms ; I1 = V/R1 = 120/576 = 0.2083 Amp hourly handyman services
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WebNow, we calculate current flowing in the bulbs ( as we know this is series circuit so same current will flow in the circuit) Current (I)= Voltage/total resistance=200/2066.67=0.0967 A Now we find the power consumed by each bulb, power consumed by 40 W bulb=I^2R= (0.0967)^2*1000=9.35 W Power consumed by 60 W bulb=I^2*R= … Web26 nov. 2024 · Let's try another problem. This time, we have two bulbs, one with a 20 volt and 200 watt power rating, and another one with 20 volt and 50 watt power rating, … Web20 jul. 2024 · We all know that the bulb which dissipates more power will glow brighter (P=I^2 * R). First, I will calculate the resistance of two bulbs. Consider 60 watts bulb P = V^2 / R R = V^2 / P = 200^2 / 60 = 666.66 ohm’s The resistance of the 60-watt bulb is 666.66 ohm’s Consider 100 watts bulb R = V^2 / P = 200 ^2 / 100 = 400 ohm’s links eye color