(a) Given a$\mathbf{48}\mathbf{.}\mathbf{0}\mathbf{-}\mathbf{v}$ battery and$\mathbf{24}\mathbf{.}\mathbf{0}\mathbf{-}\mathbf{\Omega }$ and$\mathbf{96}\mathbf{.}\mathbf{0}\mathbf{-}\mathbf{\Omega }$resistors, find the current and power for each when connected in series. (b) Repeat when the resistances are in parallel.

Electric power is the percentage of time it takes to work or generate energy and is expressed as the work W or the transmitted energy divided by the time interval t.

The voltage of the battery is $\mathrm{E}=48\mathrm{V}$

Two resistors are ${\mathrm{R}}_1=24\mathrm{\Omega }\mathrm{and}{\mathrm{R}}_2=96\mathrm{\Omega }$

(a)

If you connect resistors in series, the same current will flow through both, but the voltage will change. The total resistance of these two resistors can be expressed as,

$$\begin{gathered} {\mathrm{R}}_{\mathrm{s}}={\mathrm{R}}_1+{\mathrm{R}}_2 \\ =24\mathrm{\Omega }+96\mathrm{\Omega } \\ =120\mathrm{\Omega } \end{gathered}$$

The current that passes through this combination of resistors can be given as,

$$\begin{gathered} {\mathrm{I}}_{\mathrm{s}}=\frac{\mathrm{E}}{{\mathrm{R}}_{\mathrm{s}}} \\ =\frac{48\mathrm{V}}{120\mathrm{\Omega }} \\ =0.4\mathrm{A} \end{gathered}$$

Therefore, the power for the series combination of the resistor can be calculated as,

For

$$\begin{gathered} {\mathrm{P}}_1={{\mathrm{I}}_{\mathrm{s}}}^2{\mathrm{R}}_1 \\ ={\left(0.4\mathrm{A}\right)}^2\times \left(24\mathrm{\Omega }\right) \\ =3.84\mathrm{W} \end{gathered}$$

And

$$\begin{gathered} {\mathrm{P}}_2={{\mathrm{I}}_{\mathrm{s}}}^2{\mathrm{R}}_2 \\ ={\left(0.4\mathrm{A}\right)}^2\times \left(96\mathrm{\Omega }\right) \\ =15.36\mathrm{W} \end{gathered}$$

Therefore, the current in the series is ${\mathrm{I}}_{\mathrm{s}}=0.4\mathrm{A}$and power for each when connected in series is${\mathrm{P}}_1=3.84\mathrm{W},\mathrm{and}{\mathrm{P}}_2=15.36\mathrm{W}$

(b)

If you connect resistors in parallel, the voltage drop will be the same, but the current flowing will be different. The total resistance of these two resistors is,

$$\begin{gathered} \frac{1}{{\mathrm{R}}_{\mathrm{p}}}=\frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2} \\ \frac{1}{{\mathrm{R}}_{\mathrm{p}}}=\frac{{\mathrm{R}}_1+{\mathrm{R}}_2}{{\mathrm{R}}_1{\mathrm{R}}_2} \\ {\mathrm{R}}_{\mathrm{p}}=\frac{{\mathrm{R}}_1{\mathrm{R}}_2}{{\mathrm{R}}_1+{\mathrm{R}}_2} \end{gathered}$$

The total current going through the circuit is then

$$\begin{gathered} {\mathrm{I}}_{\mathrm{p}}=\frac{\mathrm{E}}{{\mathrm{R}}_{\mathrm{p}}} \\ =\frac{\mathrm{E}\left({\mathrm{R}}_1+{\mathrm{R}}_2\right)}{{\mathrm{R}}_1{\mathrm{R}}_2} \\ =\frac{\left(48\mathrm{V}\right)\times \left(120\mathrm{\Omega }\right)}{\left(24\mathrm{\Omega }\right)\times \left(96\mathrm{\Omega }\right)} \\ =2.5\mathrm{A} \end{gathered}$$

Now, evaluating the power for each resistor when connected in parallel,

The current going through each resistor can be obtained from the fact that they both have the same voltage as the battery, which gives,

$$\begin{gathered} {\mathrm{I}}_1=\frac{\mathrm{E}}{{\mathrm{R}}_1} \\ =\frac{48\mathrm{V}}{24\mathrm{\Omega }} \\ =2.0\mathrm{A} \\ \mathrm{and} \\ {\mathrm{I}}_2=\frac{\mathrm{E}}{{\mathrm{R}}_2} \\ =\frac{48\mathrm{V}}{96\mathrm{\Omega }} \\ =0.5\mathrm{A} \end{gathered}$$

These currents are consistent with the overall current, that is,

${\mathrm{I}}_{\mathrm{Total}}={\mathrm{I}}_1+{\mathrm{I}}_2$

The power over a resistor can be calculated as $\mathrm{P}={\mathrm{I}}^2\mathrm{R}$, therefore,

$$\begin{gathered} {\mathrm{P}}_1={\mathrm{I}}_1^2{\mathrm{R}}_1 \\ ={\left(2.0\mathrm{A}\right)}^2\times \left(24\mathrm{\Omega }\right) \\ =96.0\mathrm{W} \\ {\mathrm{P}}_2={\mathrm{I}}^2{\mathrm{R}}_2 \\ ={\left(0.5\mathrm{A}\right)}^2\times \left(96\mathrm{\Omega }\right) \\ =24\mathrm{W}. \end{gathered}$$

Hence, the current and power for resistors ${\mathrm{R}}_1\mathrm{and}{\mathrm{R}}_2$when connected in series are ${\mathrm{I}}_1=2.0\mathrm{A},\mathrm{and}{\mathrm{P}}_1=96.0\mathrm{W},{\mathrm{I}}_1=0.5\mathrm{A}\mathrm{and}{\mathrm{P}}_2=24.0\mathrm{W}$, respectively.