Question

An electric lamp of 100 ohm, a toaster of resistance 50ohm,and a water
Alter of resistance 500ohm 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

Answer:

Resistance of electric lamp (R1) = 100 Ω

Resistance of toaster (R2) = 50 Ω

Resistance of water filter (R3) = 500 Ω

Potential difference of the source (V) = 220 V

To find

(1) Resistance of an electric iron

(2) The flow of electric current through the electric iron.

The three resistors are connected in parallel as shown in the below figure.

 



 

(1) Total resistance (R) can be calculated as shown below.

1/R = 1/R₁ + 1/R₂ + 1/R₃

1/R = 1/100  + 1/50 + 1/500

1/R = ( 5 + 10 + 1 )/500

1/R = 16/500

500/16 = R

Total Resistance (R) = 500/16 Ω

Answer 2

Clarification:

As electric lamp of 100 ohm, a toaster of resistance 50 ohm,and a water alter of resistance 500 ohm are the resistances connected in parallel combination, hence resistance of an electric iron connected to the same source that takes as much current as all three appliances will be :

have to apply the formula given below:

${\underline {\boxed {\large {\bf \gray { \dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2}. . . . . . \dfrac{1}{R_n} } }}}}$

Where,

• $\sf {R_p} \to$ Resistance of electric iron.
• $\sf {R_1} \to$ Resistance of electric lamp.
• $\sf {R_2} \to$ Resistance of a toster
• $\sf {R_3} \to$ Resistance of a water alter.

Then, from this formula we can find the total resistance. After that, by the formula of ohm's law i.e V = IR, we'll calculate current through it.

______________________________

Given:

• Resitances of electric lamp, a toaster & a water alter are connected in parallel combination.

• $\sf {R_1} \to$ 100 Ω

• $\sf {R_2} \to$ 50 Ω

• $\sf {R_3} \to$ 500 Ω

• Potential difference (V) = 220 V

To calculate:

• The resistance of an electric iron connected to the same source that takes as much current as all three appliances. ( R_p )

• Current through it. ( I )

Calculation:

When the resistors are connected in parallel, the net resistance is given by,

$\star \: {\underline {\boxed {{\sf \pink { \dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2}. . . . .\dfrac{1}{R_n} } }}}} \\ \\ \\ \sf{ \longrightarrow \:\dfrac{1}{R_p} = \dfrac{1}{100} + \dfrac{1}{50} + \dfrac{1}{500} } \\ \\ \\ \sf{ \longrightarrow \:\dfrac{1}{R_p} = \dfrac{5 + 10 + 1}{500}} \\ \\ \\ \sf{ \longrightarrow \:\dfrac{1}{R_p} = \dfrac{16}{500}} \\ \\ \\ \longrightarrow \underline{\boxed{\sf{R_p = \dfrac{500}{16} }}} \: \red{\bigstar}$

Henceforth, the resistance of an electric iron is 500/16 Ω.

Also, we know that :

$\star \: {\underline {\boxed {\large {\sf \pink { V = IR} }}}} \\ \\ \sf{ \longrightarrow \: 220 = I \times \dfrac{500}{16} } \\ \\ \\ \sf{ \longrightarrow \: \frac{220}{ \cfrac{500}{16} } = I } \\ \\ \\ \sf{ \longrightarrow \: 220 \times \dfrac{16}{500} \: A = I } \\ \\ \\ \sf{ \longrightarrow \: \dfrac{352}{50} \:A \: = I } \\ \\ \\ \longrightarrow \: \underline{\boxed{\sf{7.04 \:A \: = I }}} \: \red{\bigstar}$

Henceforth, current ᴘᴀꜱꜱɪɴɢ through it is 7.04 A.