two devices of rating 44W 220V and 11W 220V r connected in series. the combination is connected across 440 mains. The fuse of which of d two device is likely to burn when switch is on? Justify.
Answers
Concept:
Here we will use the concept of ohm's law and the formula of power will be used.
So, according to the ohm's law, V=IR
Power, P=
Given:
We have two devices so we assume it as device 1 and device 2.
For device one, P1= 44W, V1= 220V
For device two, P2= 11W, V2= 220V
Supply Voltage =440V
Find:
We have to find out of two device which is likely to be burnt when we on the switch.
Solution:
Resistance of device 1, R1=
R1= =1100 ohm
Resistance of device 2, R2=
R2== 4400 ohm
Since both are connected in series
So, R= R1+R2
R= 1100+4400= 5500 ohm
Now, we calculate current in the circuit
By using ohm's law,
I= V/R= 400/5500= 0.08 A
Voltage across the device 1= IR1
= 0.08 * 1100
= 88 volt
Voltage across the device 2= IR2
= 0.08 * 4400
= 352 volt
As we know that voltage across any device greater than the rating voltage so the fuse of that device would burn. Here in our case , in second device where rating voltage is 220 V and voltage across this device is 352 V so second device will burn when switch is on.
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Answer:
The fuse of 2nd device of rating W will burn
Explanation:
According to Ohm's Law
The formula for power P = .
For first device and
For second device and
Then resistance of first device ohm
The resistance for second device ohm
Since resistance is in series ohm
By using Ohm's Law we get = A
Voltage across first device = V
Voltage across second device = V
As we know that voltage across any device greater than the rating voltage so the fuse of that device would burn.
Here in our case , in second device where rating voltage is V and voltage across this device is V so second device will burn when switch is on.