Chapter 28, Problem 30P

(a)

To determine

The current in each resistor.

Answer to Problem 30P

The current across $200\Omega$, $70.0\Omega$, $80.0\Omega$ and $20.0\Omega$ resistor is $1.00\text{A}$, $4.00\text{A}$, $3.00\text{A}$ and $8.00\text{A}$ respectively.

Explanation of Solution

The circuit diagram is shown in figure (1).

Figure-(1)

Write the expression for Kirchhoff’s voltage rule to the loop $\text{abcd}$.

    $R_1{I}_1+R_2{I}_2=V_1$                                                                                    (I)

Write the expression for Kirchhoff’s voltage rule to the loop $\text{ebcf}$.

    $R_2{I}_2+R_4{I}_4=V_2+V_1$                                                                           (II)

Write the expression for Kirchhoff’s rule for the loop $\text{eghf}$.

    $R_3{I}_3+R_4{I}_4=V_2+V_3$                                                                           (III)

Write the expression for Kirchhoff’s rule for the circuit.

    $I_4=I_1+I_2+I_3$                                                                                    (IV)

Conclusion:

Substitute $-200\Omega$ for $R_1$, $80.0\Omega$ for $R_2$ and $40.0\text{V}$ for $V_1$ in equation (I).

    $\begin{array}{c}\left(-200\Omega \right)I_1+\left(80.0\Omega \right)I_2=40.0\text{V}\\ \left(-5\Omega \right)I_1+\left(2\Omega \right)I_2=1.0\text{V}\\ \left(1\Omega \right)I_2=0.5\text{V}+\left(2.5\Omega \right)I_1\end{array}$                                           (V)

Substitute $-80\Omega$ for $R_2$, $-20\Omega$ for $R_4$, $-360\text{V}$ for $V_2$ and $-40.0\text{V}$ for $V_1$ in equation (II).

    $\begin{array}{c}\left(-80\Omega \right)I_2-\left(20\Omega \right)I_4=-360\text{V}-40.0\text{V}\\ \left(4\Omega \right)I_2+\left(1\Omega \right)I_4=20\text{V}\end{array}$                                                 (VI)

Substitute $70.0\Omega$ for $R_3$, $20.0\Omega$ for $R_4$, $360\text{V}$ for $V_2$ and $80\text{V}$ for $V_3$ in equation (III).

    $\begin{array}{c}\left(70.0\Omega \right)I_3+\left(20.0\Omega \right)I_4=360\text{V}+80\text{V}\\ \left(1\Omega \right)I_4+\left(3.5\Omega \right)I_3=22\text{V}\end{array}$                                                   (VII)

Substitute $I_1+I_2+I_3$ for $I_4$ in equation (VI).

    $\begin{array}{c}\left(4\Omega \right)I_2+\left(1\Omega \right)\left(I_1+I_2+I_3\right)=20.0\text{V}\\ \left(1\Omega \right)I_1+\left(5\Omega \right)I_2+\left(1\Omega \right)I_3=20.0\text{V}\end{array}$                                                  (VIII)

Substitute $I_1+I_2+I_3$ for $I_4$ in equation (VII).

    $\begin{array}{c}\left(1\Omega \right)\left(I_1+I_2+I_3\right)+\left(3.5\Omega \right)I_3=22\text{V}\\ \left(1\Omega \right)I_1+\left(1\Omega \right)I_2+\left(4.5\Omega \right)I_3=22\text{V}\end{array}$                                                       (IX)

Substitute $0.5\text{V}+\left(2.5\Omega \right)I_1$ for $I_2$ in equation (VIII).

    $\begin{array}{c}\left(1\Omega \right)I_1+5\left(0.5\text{V+}\left(2.5\Omega \right)I_1\right)+\left(1\Omega \right)I_3=20.0\text{V}\\ \left(13.5\Omega \right)I_1+\left(1\Omega \right)I_3=17.5\text{V}\\ \left(1\Omega \right)I_3=17.5\text{V}-\left(13.5\Omega \right)I_1\end{array}$             (X)

Substitute $0.5\text{V}+\left(2.5\Omega \right)I_1$ for $I_2$ in equation (IX).

    $\begin{array}{c}\left(1\Omega \right)I_1+\left(0.5\text{V}+\left(2.5\Omega \right)I_1\right)+\left(4.5\Omega \right)I_3=22\text{V}\\ \left(3.5\Omega \right)I_1+\left(4.5\Omega \right)I_3=21.5\text{V}\end{array}$                            (XI)

Substitute $17.5\text{V}-\left(13.5\Omega \right)I_1$ for $I_3$ in equation (XI).

    $\begin{array}{c}\left(3.5\Omega \right)I_1+\left(4.5\Omega \right)\left(17.5\text{V}-\left(13.5\Omega \right)I_1\right)=21.5\text{V}\\ \left(-57.25\Omega \right)I_1=57.25\text{V}\\ I_1=1.00\text{A}\end{array}$

Substitute $1.00\text{A}$ for $I_1$ in equation (X) to find $I_3$.

    $\begin{array}{c}\left(1\Omega \right)I_3=17.5\text{V}-\left(13.5\Omega \right)\left(1.00\text{A}\right)\\ I_3=4.00\text{A}\end{array}$

Substitute $1.00\text{A}$ for $I_1$ in equation (V) to find $I_2$.

    $\begin{array}{c}\left(1\Omega \right)I_2=0.5\text{V}+\left(2.5\Omega \right)\left(1.00\text{A}\right)\\ I_2=3.00\text{A}\end{array}$

Substitute $4.00\text{A}$ for $I_3$, $3.00\text{A}$ for $I_2$ and $1.00\text{A}$ for $I_1$ in equation (IV) to find $I_4$.

    $\begin{array}{c}{I}_4=1.00\text{A}+3.00\text{A}+4.00\text{A}\\ =8.00\text{A}\end{array}$

Therefore, the current across $200\Omega$, $70.0\Omega$, $80.0\Omega$ and $20.0\Omega$ resistor is $1.00\text{A}$, $4.00\text{A}$, $3.00\text{A}$ and $8.00\text{A}$ respectively.

(b)

To determine

The potential difference across the $200\text{-}\Omega$ resistor.

Answer to Problem 30P

The potential difference across $200\text{-}\Omega$ resistor is $200\text{V}$.

Explanation of Solution

Write the expression for potential difference across the resistor.

    $\Delta V=IR$                                                                                 (XII)

Here, $R$ is the resistance of the resistor, $\Delta V$ is the potential difference and $I$ is the current through the resistor.

Conclusion:

Substitute $200\text{-}\Omega$ for $R$ and $1.00\text{A}$ for $I$ in equation (XII) to find $\Delta V$.

    $\begin{array}{c}\Delta V=\left(1.00\text{A}\right)\left(200\text{-}\Omega \right)\\ =200\text{V}\end{array}$

Therefore, the potential difference across $200\text{-}\Omega$ resistor is $200\text{V}$.