Question

Question 2
An electric lamp of $100\Omega$, a toaster of resistance $50\Omega$, and a water filter of resistance $500\Omega$ 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

Resistance of electric lamp, $R_1=100\Omega$
Resistance of toaster, $R_2=50\Omega$
Resistance of water filter, $R_3=500\Omega$
Potential difference of the source, V = 220 V
These are connected in parallel, as shown in the following figure.

Let R be the equivalent resistance of the circuit.
$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}=\frac{1}{100}+\frac{1}{50}+\frac{1}{500}$ = $\frac{5+10+1}{500}$ = $\frac{16}{500}$
According to Ohm’s law, V = IR
$I=\frac{V}{R}$
Where,current flowing through the circuit = I
$I=\frac{220}{\frac{500}{16}}=\frac{\frac{220}{16}}{500}=7.04A$
7.04 A of current is drawn by all the three given appliances.
Therefore, current drawn by an electric iron connected to the same source of potential 220 V = 7.04 A
Let R’ be the resistance of the electric iron.
According to Ohm’s law, V = IR’
$R^{\prime }=\frac{V}{I}=\frac{220}{7.04}=31.25\Omega$
Therefore, the resistance of the electric iron is $31.25\Omega$ and the current flowing through it is 7.04 A.