Question

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

Answer 1

Step 1, Given data

Resistance of Lamp = 100 ohm

Resistance of toaster = 50 ohm

Resistance of water filter = 500 ohm

Voltage = 220 V

Step 2

We know,

From the ohm's law

V = I R

So,

$I$ in electric lamp $=\frac{220}{100}$A

$I$ in toaster $=\frac{220}{50}$A

$I$ in filter $=\frac{220}{500}$A

Now total current

$\Rightarrow$ Total current $=\frac{220}{100}+\frac{220}{50}+\frac{220}{500}$

Or, $=\frac{220}{50}(\frac{1}{2}+1+\frac{1}{10})$

Or, $=\frac{22}{5}\times \frac{8}{5}$

Or, $=\frac{176}{25}$ $=7A$

Now we can write Resistance of the iron

$\Rightarrow R$ if iron $=\frac{V}{I}=\frac{220}{\frac{176}{25}}$ $=31.25\Omega$

Hence, the resistance of the iron is $31.25\Omega .$