heat produced in a current carrying wire in 5s is 60J. the same current is passed through another wire of half the resistance .the heat produced in 5s will be a.60J b.30J c.15J d.120J
two electric bulbs have resistances in the ratio 1: 2.if they are joined in series the energy consumed in them are in the ratio
a.1:2 b 2:1 c. 4:1 .1:1
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Answered by
2
tance of the bulbs filament
+500+=+220%5E2+%2F+R+
+R+=+48400%2F500+
+R+=+96.8+
There is +110%2F2+=+55+ volts across each bulb
+55%5E2%2F96.8+=+3025%2F96.8+
+3025+%2F96.8+=+31.25+
31.25 watts is generated by each bulb
hope I got it
Answer by Alan3354(60301) About Me (Show Source):
You can put this solution on YOUR website!
Two electric bulbs each designed to operate with a 500 watts in 220 volt line are connected in series now they are connected with a 110 volt line what is the power generated by each bulb
---------------
P = E^2/R
R = E^2/P
R = 96.8 ohms
------------
In series, V across each bulb = 55 volts
----
P = 55^2/96.8
P = 31.25 watts
--------------
The bulbs don't generate any power, they consume it.
+500+=+220%5E2+%2F+R+
+R+=+48400%2F500+
+R+=+96.8+
There is +110%2F2+=+55+ volts across each bulb
+55%5E2%2F96.8+=+3025%2F96.8+
+3025+%2F96.8+=+31.25+
31.25 watts is generated by each bulb
hope I got it
Answer by Alan3354(60301) About Me (Show Source):
You can put this solution on YOUR website!
Two electric bulbs each designed to operate with a 500 watts in 220 volt line are connected in series now they are connected with a 110 volt line what is the power generated by each bulb
---------------
P = E^2/R
R = E^2/P
R = 96.8 ohms
------------
In series, V across each bulb = 55 volts
----
P = 55^2/96.8
P = 31.25 watts
--------------
The bulbs don't generate any power, they consume it.
Answered by
15
Answer:
option b 2:1co.es out to be the correct answer
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