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?

(D) Resistance of electric lamp,$R_1 = 100 \,\Omega$.
Resistance of toaster,$R_2 = 50 \,\Omega$.
Resistance of water filter,$R_3 = 500 \,\Omega$.
Voltage of the source,$V = 220 \,V$.
These are connected in parallel.
Let $R$ be the equivalent resistance of the circuit.
$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{100} + \frac{1}{50} + \frac{1}{500}$.
$\frac{1}{R} = \frac{5 + 10 + 1}{500} = \frac{16}{500}$.
$R = \frac{500}{16} \,\Omega = 31.25 \,\Omega$.
According to Ohm's law,$V = IR$,so the total current $I = \frac{V}{R}$.
$I = \frac{220}{500/16} = \frac{220 \times 16}{500} = 7.04 \,A$.
Since the electric iron takes as much current as all three appliances,the current through the iron is $I' = 7.04 \,A$.
Using Ohm's law for the electric iron,$R' = \frac{V}{I'} = \frac{220}{7.04} = 31.25 \,\Omega$.
Therefore,the resistance of the electric iron is $31.25 \,\Omega$ and the current flowing through it is $7.04 \,A$.