Question

In Figure, the ideal batteries have emfs${\mathbf{\text{ε}}}_{\mathbf{\text{1}}}\mathbf{=}\mathbf{10}\mathbf{.0}\mathbf{\text{V}}$and${\mathit{\epsilon }}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.500}\mathbf{}{\mathit{\epsilon }}_{\mathbf{1}}$ , and the resistances are each $\mathbf{4}\mathbf{.00}\mathit{\Omega }$.

(a) What is the current in resistance 2?

(b) What is the current in resistance 3?

Short answer

1. The current in resistance 2 is $0\text{A}$.
2. The current in resistance 3 is $1.25\text{A}$.

Step-by-step solution

Given data:

The voltage, ${\epsilon }_1=10.0\text{V}$

The voltage, ${\epsilon }_2=0.500{\epsilon }_1$

The resistors, $R_1=R_2=R_3=4.00\Omega$

Understanding the concept:

Apply the Kirchhoff’s loop rule and thejunction rule to each loop and solve the equations to get the current in resistance 2 and resistance 3.

Formula:

The total current is define by,

$i_3=i_1+i_2$

(a) Calculate the current in resistance 2:

Apply the loop rule to the left side closed loop,

${\epsilon }_1-i_1{R}_1-i_3{R}_3=0$ ….. (1)

But according to the junction rule,

$i_3=i_1+i_2$ ….. (2)

Substitute the above equation into equation (1).

$\begin{array}{l}{\epsilon }_1-i_1{R}_1-(i_1+i_2)R_3=0\\ {\epsilon }_1-i_1{R}_1-i_1{R}_3-i_2{R}_3=0\\ {\epsilon }_1-i_1(R_1+R_3)-i_2{R}_3=0\end{array}$

Substitute $10\text{V}$ for ${\epsilon }_1$, $4.00\Omega$for $R_1$and $R_3$, you get

role="math" localid="1662701819687" $$\begin{gathered} \begin{array}{l}10\text{V}-i_1(4.00\Omega +4.00\Omega )-i_2\left(4.00\Omega \right)=0\\ 10\text{V}-i_1\left(8.00\Omega \right)-i_2\left(4.00\Omega \right)=0\\ 10\text{V}=i_1\left(8.00\Omega \right)+i_2\left(4.00\Omega \right)\end{array} \\ \text{5V}=i_1\left(4.00\Omega \right)+i_2\left(2.00\Omega \right) \end{gathered}$$ ….. (3)

Now, applying the loop rule to the right-side closed loop,

${\epsilon }_2-i_2{R}_2-i_3{R}_3=0$ ….. (4)

But according to the junction rule,

$i_3=i_1+i_2$

Substitute the above equation into equation (4).

$\begin{array}{l}{\epsilon }_2-i_2{R}_2-(i_1+i_2)R_3=0\\ {\epsilon }_2-i_2{R}_2-i_1{R}_3-i_2{R}_3=0\\ {\epsilon }_2-i_2(R_2+R_3)-i_1{R}_3=0\end{array}$

As $5\text{V}$for ${\epsilon }_2$, $4.00\Omega$for $R_2$and $R_3$, and youhave

$$\begin{gathered} 5V-i_2(4.00\Omega +4.00\Omega )-i_1\left(4.00\Omega \right)=0 \\ 5V=i_1\left(4.00\Omega \right)+i_2\left(8.00\Omega \right) \end{gathered}$$ ….. (5)

Subtract equation(3)from equation(5),

$\begin{array}{l}5\text{V}-5\text{V}=i_1\left(4.00\Omega \right)+i_2\left(8.00\Omega \right)-i_1\left(4.00\Omega \right)-i_2\left(2.00\Omega \right)\\ 0\text{V}=i_2\left(6.00\Omega \right)\\ i_2=0\text{A}\end{array}$

Hence, the current in resistance 2 is$0\text{A}$ .

(b) Calculate the current in resistance 3:

As you have equation(3),

$\text{5V}=i_1\left(4.00\Omega \right)+i_2\left(2.00\Omega \right)$

Substitute $0\text{A}$for $i_2$ in the above equation.

$\begin{array}{c}\text{5V}=i_1\left(4.00\Omega \right)+\left(0\text{A}\right)\left(2.00\Omega \right)\\ =i_1\left(4.00\Omega \right)\end{array}$

$\begin{array}{c}{i}_1=\frac{5\text{V}}{4.00\Omega }\\ =1.25\text{A}\end{array}$

Rewrite equatopn (2) and substitute the known values.

$\begin{array}{c}{i}_3=i_1+i_2\\ =1.25\text{A}+0\text{A}\\ =1.25\text{A}\end{array}$

Hence, the current in resistance 3 is $1.25\text{A}$.