Question

An electric oven of resistance $125\Omega$, a toaster of resistance $50\Omega$ and an electric geyser of $250\Omega$ are connected in parallel to a $220V$ source. What is the resistance of an electric iron connected to the same source that takes as much as all the three appliances and what is the current through it?

Answer 1

Step 1 - Given data.

1. The resistance of the oven is $R_o=125\Omega$
2. The resistance of the toaster is $R_t=50\Omega$
3. The resistance of the geyser is $R_g=250\Omega$
4. The supplied voltage is $V=220V$

Step 2 - Finding the equivalent resistance of the parallel combination.

Consider $R_{eq}$ be the equivalent resistance of the circuit.

All the three given appliances are connected in parallel combinations.

So the equivalent resistance for the circuit is given by,

$$\begin{gathered} \frac{1}{R_{eq}}=\frac{1}{R_o}+\frac{1}{R_t}+\frac{1}{R_g} \\ \frac{1}{R_{eq}}=\frac{1}{125\Omega }+\frac{1}{50\Omega }+\frac{1}{250\Omega } \\ \frac{1}{R_{eq}}=\frac{4+10+2}{500} \\ {R}_{eq}=\frac{500}{16} \\ {R}_{eq}=31.25\Omega \end{gathered}$$

As the electric iron takes the same amount of current as the combination and is connected to the same voltage source. So its resistance will same as the equivalent resistance of the combination of those three appliances.

Step 3 - Calculate the current flow.

Assume the current flowing through the iron as $i$.

According to Ohm's law, the formula to calculate the current flow is given by,

$i=\frac{V}{R_{eq}}$.

Now substitute the values into the formula.

$$\begin{gathered} i=\frac{220V}{31.25\Omega } \\ i=7.04A \end{gathered}$$

Thus, the resistance of the electric iron is $31.25\Omega$ and the current through the iron is $7.04A$.