Science, asked by navneethsharma5046, 1 year ago

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

Answered by rajagrewal768
4

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= \frac{V^{2} }{R}

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 V_{0}=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=\frac{V1^{2} }{P1}

  R1= \frac{220^{2} }{44}=1100 ohm

Resistance of device 2, R2=\frac{V2^{2} }{P2}

R2=\frac{220^{2} }{11}= 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.

#SPJ3

Answered by syed2020ashaels
1

Answer:

The fuse of 2nd device of rating 11\\ W will burn

Explanation:

According to Ohm's Law V=IR
The formula for power P = \frac{V^{2}}{R}.
For first device P_{1} = 44W and V_{1} = 220 V
For second device P_{2} = 11 W and V_{2} = 220 V
Then resistance of first device R_{1} = \frac{220^2}{44}= 1100 ohm
The resistance for second device R_{2} = \frac{220^2}{11} = 4400 ohm
Since resistance is in series R=R_{1}+R_{2} = 1100+4400 = 5500 ohm
By using Ohm's Law we get I=\frac{V}{R} = \frac{440}{5500} = 0.08 A

Voltage across first device = IR_{1} = 0.08*1100 = 88 V
Voltage across second device = IR_{2} = 0.08 * 4400 = 352 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 220 V and voltage across this device is 352 V so second device will burn when switch is on.

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