Question

Compute the equivalent resistance of the network in Fig. E26.13, and find the current in each resistor. The battery has negligible internal resistance.

Short answer

The equivalent resistance is $5.0\Omega$ and the current in each resistor is 8.0 A, 4.0 A, 3.0 A, and 9.0 A.

Step-by-step solution

Definition of series and parallel connection

In a series circuit, all components are connected in a way that one’s tail is connected to the second’s head and so on, the current flowing in all components is the same.

In a parallel circuit, all components are connected in a way that their heads and tails are connected so that the potential drop is the same for all the components.

Determine the equivalent resistance

Given data:

• R₁ = 3
• R₂ = 6
• R₃ = 12
• R₄ = 4
• r = 0
• = 60 V

Here,$R_1$and$R_2$are in parallel, so their combination$R_{12}$can be calculated as:

$$\begin{gathered} R_{12}=\frac{R_1{R}_2}{R_1+R_2} \\ =\frac{\left(3\right)\left(6\right)}{3+6} \\ =2.0\Omega \end{gathered}$$

The same step for$R_3$and$R_4$as they are in parallel,

$$\begin{gathered} R_{34}=\frac{R_3{R}_4}{R_3+R_4} \\ =\frac{\left(12\right)\left(4\right)}{12+4} \\ =3.0\Omega \end{gathered}$$

So, the figure can be reduced as:

Now, both resistances are in series, so equivalent resistance can be given as:

$$\begin{gathered} R_{eq}=R_{12}+R_{34} \\ =2.0+3.0 \\ =5.0\Omega \end{gathered}$$

Determine the current through each resistor

Here, as$R_{12}$ and$R_{34}$ are in series, therefore, they have the same current which equals to the current flows due to the battery and from Ohm’s law, the combination can be calculated as:

$$\begin{gathered} l_{12}=l_{34} \\ =\frac{\epsilon }{R_{eq}} \\ =\frac{60.0eV}{5.0\Omega } \\ =12A \end{gathered}$$

Now,$R_1$ and$R_2$ have the same voltages as they are connected in parallel and this voltage equals to the voltage of their combination .

$V_{12}=V_1=V_2=I_{12}{R}_{12}$

Plug the values,

$$\begin{gathered} V_1=V_2 \\ =\left(12A\right)\left(2.0\Omega \right) \\ =24.0V \end{gathered}$$

Now, using Ohm’s law to calculate currents as:

$$\begin{gathered} l_1=\frac{V_1}{R_1} \\ =\frac{24.0V}{3.0\Omega } \\ =8.0A \end{gathered}$$

And

$$\begin{gathered} l_2=\frac{24.0V}{6.0\Omega } \\ =4.0A \end{gathered}$$

Similarly, we have $l_3=3.0A$and $l_4=9.0A$.

Thus, the equivalent resistance is $5.0\Omega$and the current in each resistor is 8.0 A, 4.0 A, 3.0 A, and 9.0 A.