Question

An electric heater that is connected to a $220V$ supply line has two resistance coils. A and B of $24\Omega$ resistance each. these coils can be used separately (one at a time), in series, or in parallel. Calculate the current drawn when (a) Only one coil A is used and (b) coils A and B are used in series.

Answer 1

Step 1: Given data

1. The supplied voltage, $V=220volt.$.
2. The resistance of two coils is, $24\Omega$.

Step 2: Formulae

1. We know, that current is the number of charges passing through a conductor in unit time.
2. The current flowing through a circuit of resistance is defined by the form, $I=\frac{V}{R}$, where, $V$ is the voltage across the conductor.
3. We know when two resistance $R_1$ and $R_2$ are connected in series, then the equivalent resistance of the circuit is $\left(R_1+R_2\right)$. If the two resistances are connected in parallel combination, then the equivalent resistance is $\frac{R_1{R}_2}{R_1+R_2}$.

Step 3: Current through the heater when only A resistance is connected

As we know, $I=\frac{V}{R}$

So,

$$\begin{gathered} I=\frac{V}{R}=\frac{220}{24}=9.61 \\ orI=9.61A. \end{gathered}$$

Step 4: Diagram

Step 5: Current through the heater when A and B both resistances are connected in series

Now, the equivalence resistance in this case is, $R=\left(R_1+R_2\right)$

So,

$$\begin{gathered} R=\left(24+24\right)=48 \\ orR=48\Omega \end{gathered}$$

Therefore, $I=\frac{V}{R}$

So,

$$\begin{gathered} I=\frac{V}{R}=\frac{220}{48}=4.58 \\ orI=4.58A. \end{gathered}$$

Therefore, (a) the current through the coil, when only A resistance is connected is $9.61A$ and (b) the current through the coil, when A and B both resistance are connected are series is $4.58A$.