The figure shows a section of a circuit. The resistances are $\mathbf{}{\mathit{R}}_{\mathbf{1}}\mathbf{=}\mathbf{2}\mathbf{.0}\mathit{\Omega }$ , ${\mathit{R}}_{\mathbf{2}}\mathbf{=}\mathbf{4}\mathbf{.0}\mathit{\Omega }$and ${\mathit{R}}_{\mathbf{3}}\mathbf{=}\mathbf{6}\mathbf{.0}\mathit{\Omega }$, and the indicated current is $\mathit{I}\mathbf{=}\mathbf{6}\mathbf{.0}\mathbf{\text{A}}$ . The electric potential difference between points A and B that connect the section to the rest of the circuit is ${\mathit{V}}_{\mathbf{A}}\mathbf{-}{\mathit{V}}_{\mathbf{B}}\mathbf{=}\mathbf{78}\mathbf{\text{V}}$ . (a) Is the device represented by “Box” absorbing or providing energy to the circuit, and (b) At what rate?

The resistor, $R_1=2.0\Omega$

The resistor, $R_2=4.0\Omega$

The resistor, $R_3=6.0\Omega$

The current, $I=6.0\text{A}$

The potential difference, $V_A-V_B=78\text{V}$

Using the given data, you can calculate the total power dissipated by the resistors. By comparing the total power dissipated by the resistors and the external power supplied, you can find whether the device is absorbing or providing energy.

Formula:

The Ohm’s voltage,

$V=IR$

The power is,

$P=I^{2}R$

At any junction,$\sum I=0$

The voltage across$R_3$can be calculated as,

$\begin{array}{c}{V}_3=I{R}_3\\ =6.0\text{A}\times 6.0\Omega \\ =36\text{V}\end{array}$

The voltage across$R_1$can be calculated as,

$\begin{array}{c}{V}_1=(V_A-V_B)-V_3\\ =78\text{V}-36\text{V}\\ =42\text{V}\end{array}$

$\begin{array}{c}{I}_1=\frac{V_1}{R_1}\\ =\frac{42\text{V}}{2\Omega }\\ =21\text{A}\end{array}$

Now, we apply the junction rule at point A to find the current flowing through$R_2$

$\begin{array}{c}{I}_2=I_1-I\\ =(21-6)\text{A}\\ =15\text{A}\end{array}$

Now, the total power dissipated by the resistors,

$P_{total}=P_1+P_2+P_3$

Since the power is,

$P=I^{2}R$

Therefore, the total power will becomes,

$P_{total}={I_1}^2{R}_1+{I_2}^2{R}_2+{I_3}^2{R}_3$

Substitute known values in the above equation.

$\begin{array}{c}{P}_{total}=({\left(21\right)}^2\times 2)+({\left(15\right)}^2\times 4)+({\left(6\right)}^2\times 6)\\ =882+900+216\\ =1998\text{W}\\ =1.998\text{kW}\end{array}$

But, the power supplied externally to this section will be,

$\begin{array}{c}{P}_A=I_1(V_A-V_B)\\ =21\text{A}\times 78\text{V}\\ =1638\text{A}\\ =1.638\text{kW}\end{array}$

On comparing$P_A$and$P_{total}$, as $P_{total}>P_A$ you can say that the box must is providing energy.

The rate at which the Box is supplying the energy will be,

$\begin{array}{c}{P}_{Box}=P_{total}-P_A\\ =1998-1638\\ =360\text{W}\end{array}$

Hence, the rate of the energy is .$360\text{W}$