Question

Two lamps each rated as 12 V, 36 W are connected in parallel and this combination is joined in series with a variable resistor across a 110 V supply. Find the value of the variable resistor in order that the lamps can burn at the rated wattage?

Answer 1

$\rule{300}{2}$

GIVEN :-

Two lamps each rated as 12 V, 36 W are connected in parallel and this combination is joined in series with a variable resistor across a 110 V supply.

◙ Power = 36 W

◙ Potential difference = 12 V

TO FIND :-

The value of the variable resistor in order that the lamps can burn at the rated wattage.

SOLUTION :-

We know,

◙ P = V²/R

⇒R = V²/P

⇒R = (12)²/36

⇒R = 144/36

⇒R = 4Ω

Now,

◙ Current(I) = V/R

⇒I = 12/4

⇒I = 3 A

Again,

◙ Potential difference = 110 V

◙ I = 3 A

◙ R = V/I

⇒R = 110/3

⇒R = 36.66Ω

∴ The value of the variable resistor in order that the lamps can burn at the rated wattage = 36.66Ω - 4Ω = 32.66Ω.

$\rule{300}{2}$

Check :-

Change in p.d. = 110 - 12 = 98 V

R = 32.66Ω

I = V/R = 98/32.66 = 3 A (approx.)

Answer 2

Answer:

36.6 Ω

Explanation:

Given:

• Rating of first lamp = 12 V
• Rating of second lamp = 36 W
• They are connected in series with variable resistor across a 10 volts supply

To find:

Value of variable resistor for burning at ratted wattage

Formula for power is given as: $p = \frac{v^{2} }{r}$

We have P = 36 Watts and voltage as 12 volts

Substituting the values, we get:

$36 = \frac{12^{2} }{R}$

$36 = \frac{144}{R}$

$36R = 144$

$R = 4$Ω

The value of R is 4 ohms.

As we need to find the value of variable resistor for burning at ratted wattage

V=IR

12=4I

I = 3 amperes

Current will be same and potential difference is given as 110 volts, so:

V=IR

110=3×R

R = 36.6..Ω

the value of variable resistor for burning at ratted wattage is 36.6Ω