Question

In Fig. 27-83, ${\mathit{\epsilon }}_{\mathbf{1}}\mathbf{=}\mathbf{6}\mathbf{.00}\mathbf{}\mathbf{\text{\hspace{0.17em}V}}$,${\mathit{\epsilon }}_{\mathbf{2}}\mathbf{=}\mathbf{12}\mathbf{.0}\mathbf{}\mathbf{\text{\hspace{0.17em}V}}$,${\mathit{R}}_{\mathbf{1}}\mathbf{=}\mathbf{200}\mathbf{}\mathit{\Omega }$, and${\mathit{R}}_{\mathbf{2}}\mathbf{=}\mathbf{100}\mathbf{}\mathit{\Omega }$. What are the

(a) size and

(b) direction(up or down) of the current through resistance 1, the

(c) size and

(d) direction of the current through resistance 2, and the

(e) size and

(f) direction of the current through battery 2?

Short answer

1. Magnitude of the current through resistance 1 is$i_1=0.06A$
2. The current through resistance 1 is in downward direction.
3. Magnitude of the current through resistance 2 is$i_2=0.18A$
4. The current through resistance 2 is in leftward direction.
5. Magnitude of the current through battery 2 is$i=0.24A$
6. The current through battery 2 is in upward direction.

Step-by-step solution

Determine the given quantities

Consider the value of the voltage sources:

$$\begin{gathered} {\epsilon }_1=6.00V \\ {\epsilon }_2=12.0V \end{gathered}$$

Consider the values of the resistance as:

$$\begin{gathered} R_1=200\Omega \\ {R}_2=100\Omega \end{gathered}$$

Determine the concept Kirchhoff’s Laws

Consider the concept of Kirchhoff’s voltage law and current law to find the unknown currents and their direction. By applying KVL to both loops, determine two different equations.

Consider the formulas:

According to KVL,$\sum V=0$

According to KCL,$\sum I_{in}=\sum I_{out}$

Step 3:(a)Determine the magnitude of current i1

Applying KVL to first loop:

${\epsilon }_1+i_1{R}_1-i_2{R}_2=0$…… (1)

Applying KVL to second loop:

${\epsilon }_2-i_1{R}_1=0$

Hence, to find current$i_1:$

$i_1=\frac{{\epsilon }_2}{R_1}$

Substitute the values and solve as:

$\begin{array}{c}{i}_1=\frac{12}{200}\\ =0.06A\end{array}$

Size of the current through resistance 1 is$i_1=0.06A$

Step 4:(b) Determine the direction of the current i1

Current$\left(i_1\right)$is directed in downward direction to complete the one cycle.

Therefore, the direction of the current is in the downward direction.

Step 5:(c) Determine the magnitude of current i2

Substitute the valuesof$i_1$ in equation.$\left(1\right)$

$6.00+(0.06\times 200)-(i_2\times 100)=0$

Hence, the value of the current is:

$\begin{array}{c}{i}_2=\frac{18.0}{100}\\ =0.18A\end{array}$

Size of the current through resistance 2 is$i_2=0.18A$

Step 6:(d) Solve for direction of the current i2

Current $i_2$is directed in leftward direction as suggested by the positive value of the current through the resistance $R_2$.

Step 7:(e) Determine the size of current passing through battery 2

Current through the battery2

$i=i_1+i_2$

Substitute the values and solve as:

$\begin{array}{c}i=0.06+0.18\\ =0.24A\end{array}$

Size of the current through battery 2 is $i=0.24A$.

:(f) Solve for direction of the current i

The current through battery 2 is directed in upward direction.