Question

29.
Two devices of rating 22 W; 220 V and 11 W;220 V are connected in series. The combination is
connected across a 440 V mains. The fuse of which of the two devices is likely to burn when
switch is on ?​

Answer 1

Given

• Device 1 :

● Power = 22 W

● Voltage = 220 V

• Device 2 :

● Power = 11 W

● Voltage = 220 V

To Find

• The device which consumes more current

Solution

☯ P = V²/R

☯ I ∝ 1/R

━━━━━━━━━━━━━━━━━━━━━━━━━

✭ According to the Question :

For Device 1,

→ P = V²/R

→ 22 = 220²/R

→ R = 48400/22

→ R = 2200 Ω

For Device 2,

→ P = V²/R

→ 11 = 220²/R

→ R = 48400/11

→ R = 4400 Ω

━━━━━━━━━━━━━━━━━━━━━━━━━

So, the device with the highest resistance would consume lesser current as I (current) is inversely proportional to R (resistance)

• The device with P = 22 W would more likely break the fuse (Device 1)

Answer 2

Given :-

• Two devices of rating 22 W ; 220 V and 11 W ; 220 V are connected in series. The combination is connected accross a 440 V mains.

To Find :-

• Which of the two devices is likely to burn when switch is on.

Formula Used :-

$\longmapsto \sf\boxed{\bold{\pink{P =\: \dfrac{{V}^{2}}{R}}}}$

where,

• P = Power
• V = Voltage
• R = Resistance

Solution :-

${\large{\bold{\purple{\underline{\leadsto\: In\: case\: of\: the\: first\: device\: :-}}}}}$

Given :-

• Power = 22 W
• Voltage = 220 V

According to the question by using the formula we get,

$\implies \sf 22 =\: \dfrac{{(220)}^{2}}{R}$

$\implies \sf 22 =\: \dfrac{220 \times 220}{R}$

$\implies \sf 22 =\: \dfrac{48400}{R}$

By doing cross multiplication we get,

$\implies \sf 22R =\: 48400$

$\implies \sf R =\: \dfrac{\cancel{48400}}{\cancel{22}}$

$\implies \sf\bold{\red{R =\: 2200\: \Omega}}$

Again,

${\large{\bold{\purple{\underline{\leadsto\: In\: case\: of\: the\: second\: device\: :-}}}}}$

Given :

• Power = 11 W
• Voltage = 220 V

According to the question by using the formula we get,

$\implies \sf 11 =\: \dfrac{{(220)}^{2}}{R}$

$\implies \sf 11 =\: \dfrac{220 \times 220}{R}$

$\implies \sf 11 =\: \dfrac{48400}{R}$

By doing cross multiplication we get,

$\implies \sf 11R =\: 48400$

$\implies \sf R =\: \dfrac{\cancel{48400}}{\cancel{11}}$

$\implies \sf\bold{\red{R =\: 4400\: \Omega}}$

Now, as we know that the highest resistance would be consumed less current and it is inversely proportional to R.

So, the device one will be burn when switch is on.

$\therefore$ The device one is likely to burn when switch is on.