Question

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.

Answer 1

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.

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Answer 2

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.