Request a call back

Join NOW to get access to exclusive study material for best results

CBSE Class 10 Answered

An electric bulb rated for 200w and 100v is connected to a circuit having a 200v supply. What is the resistance R that must be put in series with the bulb so that the bulb draws 200w?
Asked by saradamahapatra9999 | 07 Mar, 2019, 10:29: AM
answered-by-expert Expert Answer
Rated power of the bulb is 200 W and operating voltage is 100 V.
 
Current drawn by the bulb = power/voltage = 200/100 = 2 A
 
Resistance = Voltage/current = 100/2 = 50 Ω
 
when this bulb is connected in 200 V supply, we need to ensure the current passing through this bulb is 2 A
by connecting additional resistance.
 
total resistance required to get 2 A from 200 V supply = 200/2 = 100 Ω
 
Since resistance of bulb is 50 Ω , we need to add additional 50 Ω resistance in series.
Answered by Thiyagarajan K | 07 Mar, 2019, 11:36: AM
CBSE 10 - Physics
Asked by shrilakshmimunoli | 01 Mar, 2024, 01:15: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 10 - Physics
Asked by khajannirwan | 27 Feb, 2024, 10:20: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 10 - Physics
Asked by sailakshmi.avinesh | 13 Feb, 2024, 07:03: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 10 - Physics
Asked by sakshikatewa0 | 08 Feb, 2024, 12:48: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 10 - Physics
Asked by saanviyadla | 24 Jan, 2024, 07:06: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 10 - Physics
Asked by kamalaranjanmohantymohanty5 | 06 Jan, 2024, 10:05: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 10 - Physics
Asked by saritavishwakarma1986 | 04 Jan, 2024, 12:23: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
Get Latest Study Material for Academic year 24-25 Click here
×