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?

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Solution

## 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) $V={I}_{2}{R}_{eq}\phantom{\rule{0ex}{0ex}}{I}_{2}=\frac{V}{{R}_{eq}}=\frac{220}{48}=4.58A$ Case C: If resistances are in parallelLet ${R}_{eq2}$be the equivalent resistance in parallel and ${I}_{3}$ be the current when resistances are connected in parallel. $\frac{1}{{R}_{eq2}}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}\phantom{\rule{0ex}{0ex}}\frac{1}{{R}_{eq2}}=\frac{1}{24}+\frac{1}{24}\phantom{\rule{0ex}{0ex}}\frac{1}{{R}_{eq2}}=\frac{2}{24}\phantom{\rule{0ex}{0ex}}\frac{1}{{R}_{eq2}}=\frac{1}{12}$ ${R}_{eq2}=12\Omega$ 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, ${I}_{1}=9.166A\phantom{\rule{0ex}{0ex}}{I}_{2}=4.58A\phantom{\rule{0ex}{0ex}}{I}_{3}=18.33A$.

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