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

The resistance of the electric lamp is $R_l=100\Omega$

The resistance of the toaster is $R_t=50\Omega$

The resistance of the water filter is $R_f=500\Omega$

Supplied voltage is $V=220volt.$

Step 2: Ohm's law

1. Ohm's law states that the current flowing through a conductor is directly proportional to the potential difference across the conductor.
2. The mathematical form of Ohm's law is defined by the form, $V=IR$, where, V is the potential difference, I is the current flowing through the circuit and R is the resistance of the conductor.

Step 3: Diagram

Step 4: Calculating electric lamp current

From ohm's law, we know, $V=IR$

So, the current through the electric lamp is,

$$\begin{gathered} I_L=\frac{V}{R}=\frac{220}{100}=2.2 \\ or{I}_L=2.2A. \end{gathered}$$

The current through the toaster is,

$$\begin{gathered} I_t=\frac{V}{R}=\frac{220}{50}=4.4 \\ or{I}_t=4.4A. \end{gathered}$$

Similarly, the current through the water filter is,

$$\begin{gathered} I_L=\frac{V}{R}=\frac{220}{500}=0.44 \\ or{I}_L=0.44A. \end{gathered}$$

From the question, the current flowing through the electric icon is equivalent to the sum of the current through the electric lamp, toaster, and water filter.

So, the current through the electric icon is

$$\begin{gathered} I=I_L+I_f+I_t=2.2+4.4+0.44 \\ orI=7.04A \end{gathered}$$

Step 5: Calculating the resistance of the electric iron

Now the resistance to the iron is,

$$\begin{gathered} R=\frac{V}{I}=\frac{220}{7.04}=32.25 \\ orR=32.25ohm. \end{gathered}$$

Therefore, the resistance of an electric iron is $32.25ohm$ and current through is $7.04A.$