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Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5A?
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Answers
Answered by
1
Maximum power the power supply can deliver = Supply Voltage Х Maximum current = (220 Х 5) W
Power consumed by ‘n’ bulbs = (n Х 10) W
So, n Х 10 = 220 Х 5
Or, n = 110
Thus, 110 bulbs are to b
Power consumed by ‘n’ bulbs = (n Х 10) W
So, n Х 10 = 220 Х 5
Or, n = 110
Thus, 110 bulbs are to b
Answered by
3
Resistance R1 of the bulb is given by the expression,
Supply voltage, V = 220 V
Maximum allowable current, I = 5 A
Rating of an electric bulb P=10watts
Because R=V2/P
R1=(220)2÷10
=4840
According to Ohms law,
1/R = 1/R1+ 1/R2+........ up to x times
1/R = 1/R2× x
x= R1/R
V = I R
Let R is the total resistance of the circuit for x number of electric bulbs
R=V/I
=220/5=44
therefore x=R1/R
=4840/44
=110
Resistance of each electric bulb, R1=4840
Number of electric bulbs connected in parallel are 110.
Supply voltage, V = 220 V
Maximum allowable current, I = 5 A
Rating of an electric bulb P=10watts
Because R=V2/P
R1=(220)2÷10
=4840
According to Ohms law,
1/R = 1/R1+ 1/R2+........ up to x times
1/R = 1/R2× x
x= R1/R
V = I R
Let R is the total resistance of the circuit for x number of electric bulbs
R=V/I
=220/5=44
therefore x=R1/R
=4840/44
=110
Resistance of each electric bulb, R1=4840
Number of electric bulbs connected in parallel are 110.
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