Chapter 28, Problem 60AP

(a)

To determine

The equivalent resistance of the circuit given.

Answer to Problem 60AP

The equivalent resistor of the circuit is $15.00\Omega$.

Explanation of Solution

Following figure shows the resistance connected a voltage source of $15.0\text{V}$ in closed circuit.

Figure-(1)

Write the expression for equivalent resistance connected in parallel.

    $R_{\text{peq}}=\frac{R_1{R}_2}{R_1+R_2}$                                                                                                            (I)

Here, $R_{\text{peq}}$ is the equivalent resistance connected in parallel, $R_1$ and $R_2$ are the magnitude of the resistors.

Write the expression for equivalent resistance connected in series.

    $R_{\text{seq}}=R_1+R_2+R_3$                                                                                                     (II)

Here, $R_{\text{seq}}$ is the equivalent resistance connected in series.

Following figure shows the rearranged circuit of the system from Figure (1) for better understanding.

Figure-(2)

Conclusion:

From figure (2) calculate the equivalent resistance for $9.00\Omega$ and $6.00\Omega$ resistor connected in parallel.

Substitute $6.00\Omega$ for $R_1$ and $9.00\Omega$ for $R_2$ in equation (I) to calculate $R_{\text{peq}}$.

    $\begin{array}{c}{R}_{\text{peq}}=\frac{\left(6.00\Omega \right)\left(9.00\Omega \right)}{6.00\Omega +9.00\Omega }\\ =\frac{54.00\Omega }{15.00\Omega }\\ =3.60\Omega \end{array}$

Following figure shows the reduced circuit for the above equivalence.

Figure-(3)

From figure (3) calculate the equivalent resistance for $3.60\Omega$ and $2.40\Omega$ resistor connected in series.

Substitute $3.60\Omega$ for $R_1$, $2.40\Omega$ for $R_2$ and $0.00\Omega$ for $R_3$ in equation (II) to calculate $R_{\text{seq}}$.

    $\begin{array}{c}{R}_{\text{seq}}=3.60\Omega +2.40\Omega \\ =6.00\Omega \end{array}$

Following figure shows the reduced circuit for the above equivalence.

Figure-(4)

From figure (4) calculate the equivalent resistance for $6.00\Omega$ and $6.00\Omega$ resistor connected in parallel.

Substitute $6.00\Omega$ for $R_1$ and $6.00\Omega$ for $R_2$ in equation (I) to calculate $R_{\text{peq}}$.

    $\begin{array}{c}{R}_{\text{peq}}=\frac{\left(6.00\Omega \right)\left(6.00\Omega \right)}{6.00\Omega +6.00\Omega }\\ =\frac{36.00\Omega }{12.00\Omega }\\ =3.00\Omega \end{array}$

Following figure shows the reduced circuit for the above equivalence.

Figure-(5)

From figure (5) calculate the equivalent resistance for $3.00\Omega$, $6.00\Omega$ and $6.00\Omega$ resistor connected in series.

Substitute $3.00\Omega$ for $R_1$, $6.00\Omega$ for $R_2$ and $6.00\Omega$ for $R_3$ in equation (II) to calculate $R_{\text{eq}}$.

    $\begin{array}{c}{R}_{\text{eq}}=3.00\Omega +6.00\Omega +6.00\Omega \\ =15.00\Omega \end{array}$

Therefore the equivalent resistor for the circuit is $15.00\Omega$.

(b)

To determine

The potential difference across each resistor.

Answer to Problem 60AP

voltage between junctions $a$ and $c$ is $6.00\text{V}$, $c$ and $d$ is $3.00\text{V}$, $c$ and $e$ is $1.2\text{V}$, $e$ and $d$ is $1.8\text{V}$, $f$ and $d$ is $1.8\text{V}$ and $d$ and $b$ is $6.00\text{V}$.

Explanation of Solution

Write the expression for potential difference between two points of a resistor.

    $V=IR$                                                                                                                    (III)

Here, $V$ is the potential difference between two points of a resistor, $I$ is the current flowing through the circuit and $R$ is the resistor.

Conclusion:

Substitute $15.00\Omega$ for $R$ and $15.0\text{V}$ for $V$ in equation (III) to calculate $I$.

    $\begin{array}{c}15.0\text{V}=\left(I\right)\left(15.00\Omega \right)\\ I=\frac{15.0\text{V}}{15.00\Omega }\\ I=1.00\text{A}\end{array}$

Calculate voltage drop across each $6.00\Omega$ resistor from figure (5).

Substitute $1.00\text{A}$ for $I$ and $6.00\Omega$ for $R$ in equation (III) to calculate $V_{\text{ac}}$ and $V_{\text{bd}}$.

    $\begin{array}{c}{V}_{\text{ac}}=V_{\text{bd}}\\ =\left(1.00\text{A}\right)\left(6.00\Omega \right)\\ =6.00\text{V}\end{array}$

Calculate voltage drop across $3.00\Omega$ resistor from figure (5).

Substitute $1.00\text{A}$ for $I$ and $3.00\Omega$ for $R$ in equation (III) to calculate $V_{\text{cd}}$

    $\begin{array}{c}{V}_{\text{cd}}=\left(1.00\text{A}\right)\left(3.00\Omega \right)\\ =3.00\text{V}\end{array}$

Calculate current through $2.40\Omega$ resistor.

Substitute $6.00\Omega$ for $R$ and $3.00\text{V}$ for $V$ in Equation (III) to calculate $I_{\text{ce}}$.

    $\begin{array}{c}3.00\text{V}=I_{\text{ce}}\left(6.00\Omega \right)\\ I_{\text{ce}}=\frac{3.00\text{V}}{6.00\Omega }\\ =0.5\text{A}\end{array}$

Calculate voltage drop across $2.40\Omega$ resistor.

Substitute $0.5\text{A}$ for $I$ and $2.40\Omega$ for $R$ in equation (III) to calculate $V_{\text{ce}}$

    $\begin{array}{c}{V}_{\text{ce}}=\left(0.5\text{A}\right)\left(2.40\Omega \right)\\ =1.2\text{V}\end{array}$

Calculate the voltage drop across $3.60\Omega$ resistor from figure (3).

Substitute $3.60\Omega$ for $R$ and $0.5\text{A}$ for $I$ in Equation (III) to calculate $V_{\text{ed}}$

    $\begin{array}{c}{V}_{\text{ed}}=\left(3.60\Omega \right)\left(0.5\text{A}\right)\\ =1.8\text{V}\end{array}$

Therefore, voltage drop across the parallel resistor $9.00\Omega$ and $6.00\Omega$ from Figure (2) is $1.8\text{V}$.

Therefore, voltage between junctions $a$ and $c$ is $6.00\text{V}$, $c$ and $d$ is $3.00\text{V}$, $c$ and $e$ is $1.2\text{V}$, $e$ and $d$ is $1.8\text{V}$, $f$ and $d$ is $1.8\text{V}$ and $d$ and $b$ is $6.00\text{V}$.

(c)

To determine

The each current indicated in the circuit.

Answer to Problem 60AP

ThecCurrent $I_1$ is $1.00\text{A}$, $I_2$ is $0.5\text{A}$, $I_3$ is $0.5\text{A}$, $I_4$ is $0.3\text{A}$ and $I_5$ is $0.2\text{A}$.

Explanation of Solution

Rewrite Equation (III) to calculate $I$.

    $\begin{array}{c}V=IR\\ I=\frac{V}{R}\end{array}$                                                                                                                    (IV)

Conclusion:

Substitute $6.00\Omega$ for $R$ and $6.00\text{V}$ for $V$ in Equation (IV) to calculate $I_1$.

    $\begin{array}{c}{I}_1=\frac{6.00\text{V}}{6.00\Omega }\\ =1.00\text{A}\end{array}$

Substitute $6.00\Omega$ for $R$ and $3.00\text{V}$ for $V$ in Equation (IV) to calculate $I_2$.

    $\begin{array}{c}{I}_2=\frac{3.00\text{V}}{6.00\Omega }\\ =0.5\text{A}\end{array}$

Substitute $2.40\Omega$ for $R$ and $1.2\text{V}$ for $V$ in Equation (IV) to calculate $I_3$.

    $\begin{array}{c}{I}_3=\frac{1.2\text{V}}{2.40\Omega }\\ =0.5\text{A}\end{array}$

Substitute $6.00\Omega$ for $R$ and $1.8\text{V}$ for $V$ in Equation (IV) to calculate $I_4$.

    $\begin{array}{c}{I}_4=\frac{1.8\text{V}}{6.00\Omega }\\ =0.3\text{A}\end{array}$

Substitute $9.00\Omega$ for $R$ and $1.8\text{V}$ for $V$ in Equation (IV) to calculate $I_5$.

    $\begin{array}{c}{I}_5=\frac{1.8\text{V}}{9.00\Omega }\\ =0.2\text{A}\end{array}$

Therefore, current $I_1$ is $1.00\text{A}$, $I_2$ is $0.5\text{A}$, $I_3$ is $0.5\text{A}$, $I_4$ is $0.3\text{A}$ and $I_5$ is $0.2\text{A}$.

(d)

To determine

The power delivered to each resistor.

Answer to Problem 60AP

Power delivered to resistor $R_1$ is $6.00\text{W}$, $R_2$ is $1.50\text{W}$, $R_3$ is $6.00\text{W}$, $R_4$ is $0.60\text{W}$, $R_5$ is $0.54\text{W}$ and $R_6$ is $0.36\text{W}$.

Explanation of Solution

Write the expression for power in resistor.

    $P=IV$                                                                                                                      (V)

Here, $P$ is the power delivered in the resistor.

Conclusion:

Substitute $6.00\Omega$ for $R$ and $1.00\text{A}$ for $I$ in Equation (V) to calculate $P_1$.

    $\begin{array}{c}{P}_1=\left(6.00\Omega \right)\left(1.00\text{A}\right)\\ =6.00\text{W}\end{array}$

Substitute $6.00\Omega$ for $R$ and $0.50\text{A}$ for $I$ in Equation (V) to calculate $P_2$.

    $\begin{array}{c}{P}_2=\left(6.00\Omega \right)\left(0.50\text{A}\right)\\ =1.50\text{W}\end{array}$

Substitute $6.00\Omega$ for $R$ and $1.00\text{A}$ for $I$ in Equation (V) to calculate $P_3$.

    $\begin{array}{c}{P}_3=\left(6.00\Omega \right)\left(1.00\text{A}\right)\\ =6.00\text{W}\end{array}$

Substitute $2.40\Omega$ for $R$ and $0.50\text{A}$ for $I$ in Equation (V) to calculate $P_4$.

    $\begin{array}{c}{P}_4=\left(2.40\Omega \right)\left(0.50\text{A}\right)\\ =0.60\text{W}\end{array}$

Substitute $6.00\Omega$ for $R$ and $0.30\text{A}$ for $I$ in Equation (V) to calculate $P_5$.

    $\begin{array}{c}{P}_5=\left(6.00\Omega \right)\left(0.30\text{A}\right)\\ =0.54\text{W}\end{array}$

Substitute $9.00\Omega$ for $R$ and $0.20\text{A}$ for $I$ in Equation (V) to calculate $P_6$.

    $\begin{array}{c}{P}_6=\left(9.00\Omega \right)\left(0.20\text{A}\right)\\ =0.36\text{W}\end{array}$

Therefore, power at resistor $R_1$ is $6.00\text{W}$, $R_2$ is $1.50\text{W}$, $R_3$ is $6.00\text{W}$, $R_4$ is $0.60\text{W}$, $R_5$ is $0.54\text{W}$ and $R_6$ is $0.36\text{W}$.