Question

An electric lamp of $100\mathrm{Ohms}$, a toaster of resistance of $50\mathrm{Ohms}$, and a water filter of resistance of $500\mathrm{Ohms}$ are connected in parallel to a $220\mathrm{V}$ 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

Step 1: Given information

Resistance of electric lamp (R₁) = $100\mathrm{\Omega }$

Resistance of toaster (R₂) = $50\mathrm{\Omega }$

Resistance of water filter (R₃) = $500\mathrm{\Omega }$

A potential difference of the source (V) = $220\mathrm{V}$

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

Step 2: Finding the resistance of an electron iron

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

$\frac{1}{\mathrm{R}}=\frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2}+\frac{1}{{\mathrm{R}}_3}$

By substituting the given values, we get

$$\begin{gathered} \frac{1}{\mathrm{R}}=\frac{1}{100}+\frac{1}{50}+\frac{1}{500} \\ =\frac{5+10+1}{500} \\ =\frac{16}{500} \end{gathered}$$

By applying cross-multiplication, we get

$$\begin{gathered} \mathrm{R}=\frac{500}{16}\mathrm{\Omega } \\ \mathrm{R}=31.25\mathrm{\Omega } \end{gathered}$$

Step 3: Finding the electric current through the electric iron

According to the Ohm's law,

$\mathrm{V}\propto \mathrm{I}\mathrm{or}\mathrm{V}=\mathrm{IR}$

Now,

$\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}$

By substituting the values, we get

$$\begin{gathered} \frac{500}{16}=\frac{220}{\mathrm{I}} \\ \mathrm{I}=220\times \frac{16}{500} \\ \mathrm{I}=7.04\mathrm{A} \end{gathered}$$

Therefore, the resistance is $31.25\mathrm{\Omega }$ and the current is $7.04\mathrm{A}$.