Question

A hot plate of an electric oven connected to a 220 $V$ line has two resistance coils A and B, each of 24 resistances, which may be used separately, in series, or in parallel. What are the currents in the three cases?

Answer 1

Step1: Given data-

Voltage $V$=$220V$

Resistance of coil A($R_1$)=$24\Omega$

Resistance of coil B($R_2$)=$24\Omega$

Step 2: Formula used

$V=IR$

Step 3: Calculating the currents in different situations:

Case A: If resistance is used, individual

Let $I_1$ be the current when resistance used separately.

According to ohm's law $V=I_1{R}_1$ (V=Voltage.I=current.R=Resistance)

$I_1=\frac{V}{R_1}=\frac{220}{24}=9.166A$

Case B: If resistances are in series

Let $R_{eq}$be the equivalent resistance in series and $I_2$ be the current when resistances are connected in series. $\begin{array}{l}{R}_{eq}=R_1+R_2\\ =24+24=48\Omega \end{array}$

According to ohm's law $V=I_2{R}_{eq}$ (V=Voltage.I=current.R=Equivalent Resistance) $$\begin{gathered} V=I_2{R}_{eq} \\ {I}_2=\frac{V}{R_{eq}}=\frac{220}{48}=4.58A \end{gathered}$$

Case C: If resistances are in parallel

Let $R_{eq2}$be the equivalent resistance in parallel and $I_3$ be the current when resistances are connected in parallel. $$\begin{gathered} \frac{1}{R_{eq2}}=\frac{1}{R_1}+\frac{1}{R_2} \\ \frac{1}{R_{eq2}}=\frac{1}{24}+\frac{1}{24} \\ \frac{1}{R_{eq2}}=\frac{2}{24} \\ \frac{1}{R_{eq2}}=\frac{1}{12} \end{gathered}$$

$R_{eq2}=12\Omega$

According to ohm's law $V=I_3{R}_{eq2}$ (V=Voltage.I=current.R=Equivalent Resistance) $I_3=\frac{V}{R}=\frac{220}{12}=18.33A$

Thus, $$\begin{gathered} I_1=9.166A \\ {I}_2=4.58A \\ {I}_3=18.33A \end{gathered}$$.