Question

An electric lamp of 100, a toaster of resistance 50, and a water filter of resistance 500 are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?​

Answer 1

Correct question :-

An electric lamp of 100Ω, a toaster of resistance 50Ω, and a water filter of resistance 500Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?

AnSwer :-

Given :-

☄ Resistance of electric lamp, R₁ = 100Ω

☄ Resistance of the toaster, R₂ = 50Ω

☄ Resistance of the water filter, R$_3$ = 500Ω

☄ Potential difference, V = 220V.

To Find :-

(a) the resistance of an electric iron

(b) the current passing through electric iron

Solution :-

(a) The combined resistance R of there resistor or electric devices R₁, R₂ and R$_3$, are connected in parallel

We know that,

the parallel combined resistance R is given by the formula, .i.e.,

${ \bf {\leadsto{\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}}}}$

${ \sf{\leadsto{\dfrac{1}{R} = \dfrac{1}{100} + \dfrac{1}{50} + \dfrac{1}{500}}}}$

${ \sf{\leadsto{\dfrac{1}{R} = \dfrac{5 + 10 + 1}{500} }}}$

${ \sf{\leadsto{\dfrac{1}{R} = \dfrac{16}{500} }}}$

${ \sf{\leadsto{R = \dfrac{500}{16} = \dfrac{125}{4} }}}$

thus, the resistance of electric iron is 125/4 .i.e., 31.25 Ω.

(b) using ohms law .i.e.,

$\bigstar$The potential difference, V across the ends of the given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided it's temperature remains constant the same.

In other words,

${ \leadsto{\bf I = \dfrac{V}{R}}}$

${ \leadsto{\sf I = \dfrac{220}{125/4}}}$

${ \leadsto{\sf I = \dfrac{220 \times 4}{125}}}$

${ \leadsto{\sf I = 7.04A}}$

thus, the current passing through the electric iron is 7.04 amperes

Answer 2

Resistance of electric lamp, R₁ = 100Ω

☄ Resistance of the toaster, R₂ = 50Ω

☄ Resistance of the water filter, R_3

3

= 500Ω

☄ Potential difference, V = 220V.

To Find :-

(a) the resistance of an electric iron

(b) the current passing through electric iron

Solution :-

(a) The combined resistance R of there resistor or electric devices R₁, R₂ and R_3

3

, are connected in parallel

We know that,

the parallel combined resistance R is given by the formula, .i.e.,

{ \bf {\leadsto{\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}}}}⇝

R

1

=

R

1

1

+

R

2

1

+

R

3

1

{ \sf{\leadsto{\dfrac{1}{R} = \dfrac{1}{100} + \dfrac{1}{50} + \dfrac{1}{500}}}}⇝

R

1

=

100

1

+

50

1

+

500

1

{ \sf{\leadsto{\dfrac{1}{R} = \dfrac{5 + 10 + 1}{500} }}}⇝

R

1

=

500

5+10+1

{ \sf{\leadsto{\dfrac{1}{R} = \dfrac{16}{500} }}}⇝

R

1

=

500

16

{ \sf{\leadsto{R = \dfrac{500}{16} = \dfrac{125}{4} }}}⇝R=

16

500

=

4

125

thus, the resistance of electric iron is 125/4 .i.e., 31.25 Ω.

(b) using ohms law .i.e.,

\bigstar★ The potential difference, V across the ends of the given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided it's temperature remains constant the same.

In other words,

{ \leadsto{\bf I = \dfrac{V}{R}}}⇝I=

R

V

{ \leadsto{\sf I = \dfrac{220}{125/4}}}⇝I=

125/4

220

{ \leadsto{\sf I = \dfrac{220 \times 4}{125}}}⇝I=

125

220×4

{ \leadsto{\sf I = 7.04A}}⇝I=7.04A

thus, the current passing through the electric iron is 7.04 amperes