Chapter 28, Problem 54AP

(a)

To determine

The current in the $5.00\text{Ω}$ capacitor.

Answer to Problem 54AP

The current in $5.00\text{Ω}$ capacitor is $0.7058\text{A}$.

Explanation of Solution

Write the expression based on the junction rule.

    ${\displaystyle \sum _{\text{junction}}I}=0$

Here, ${\displaystyle \sum _{\text{junction}}I}$ is the sum of currents at the junction.

Write the expression to obtain the loop rule.

    ${\displaystyle \sum _{\text{closed}\text{loop}}\Delta V}=0$

Here, ${\displaystyle \sum _{\text{closed}\text{loop}}\Delta V}$ is the sum of the potential drop across each element in a closed circuit.

Write the expression for the potential difference based on Ohm’s law.

    $V=IR$

Here, $V$ is the voltage, $I$ is the current and $R$ is the resistance.

The flow of current in the circuit is as shown in the figure below.

Figure-(1)

Here, $i_1$ and $i_2$ are the current in different sections of the circuit.

Write the equation of Kirchhoff’s voltage rule in the left loop.

    $\begin{array}{c}5i_1+\left(i_1-i_2\right)8=0\\ 5i_1+8i_1-8i_2=0\\ 13i_1-8i_2=0\end{array}$                                                                                                       (I)

Write the equation of Kirchhoff’s voltage rule in the right loop.

    $\begin{array}{c}-15+\left(i_2-i_1\right)8+10i_2=0\\ 8i_2-8i_1+10i_2=15\\ -8i_1+18i_2=15\end{array}$                                                                                        (II)

Conclusion:

Solve equation (I) and (II) to calculate $i_1$ and $i_2$.

    $\begin{array}{l}{i}_1=0.7058\text{A}\\ i_2=1.1470\text{A}\end{array}$

Therefore, the current in $5.00\text{Ω}$ capacitor is $0.7058\text{A}$.

(b)

To determine

The power delivered to the $5.00\text{Ω}$ resistor.

Answer to Problem 54AP

The power delivered to the $5.00\text{Ω}$ resistor is $2.5\text{W}$.

Explanation of Solution

Write the expression to obtain the power across the $5.00\Omega$ resistor.

    $P=i_1^2{R}_{5\text{Ω}}$

Here, $P$ is the power across the $5.00\text{Ω}$ resistor, $i_1$ is the current across the $5.00\text{Ω}$ resistor $R_{5\text{Ω}}$ is the $5\text{Ω}$ resistance.

Conclusion:

Substitute $0.7058\text{A}$ for $i_1$ and $5.00\text{Ω}$ for $R_{5\text{Ω}}$ in the above equation to calculate $P$.

    $\begin{array}{c}P={\left(0.7058\text{A}\right)}^2\left(5.00\text{Ω}\right)\\ =\left(0.4981\text{A}\right)\left(5.00\text{Ω}\right)\\ =2.5\text{W}\end{array}$

Therefore, the power delivered to the $5.00\text{Ω}$ resistor is $2.5\text{W}$.

(c)

To determine

The circuit in which Kirchhoff’s rule is required in find the value of current.

Answer to Problem 54AP

The Kirchhoff’s rule is required to find the value of current in case of circuit (c).

Explanation of Solution

In case of circuit (c), the current is flowing across each resistor and voltage drops at each resistor. As both the batteries of same emf are not in the same loop, thus, some amount of current will flow in the circuit. Hence, Kirchhoff’s rule is applicable in this case.

Conclusion:

In the case of the other two circuits, both the batteries of same emf are in the same loop. Thus they cancel out each other and no current flow in the circuit (b) and circuit (d).

Therefore, The Kirchhoff’s rule is required in find the value of current in case of circuit (c).

(d)

To determine

The circuit in which the smallest amount of power is delivered to the $10\text{Ω}$ resistor.

Answer to Problem 54AP

The smallest power is delivered across the $10.0\text{Ω}$ resistor in case of circuit (c) which is equal to $34.33\text{W}$ as in other two cases power across the $10.0\text{Ω}$ is zero.

Explanation of Solution

Conclusion:

The current flow diagram for circuit (b) is as shown in the figure below.

Figure-(2)

Write the equation of Kirchhoff’s voltage rule in right loop.

    $\begin{array}{c}15-10i_2+\left(i_1-i_2\right)8-15=0\\ -10i_2+\left(i_1-i_2\right)8=0\\ -10i_2+8i_1-8i_2=0\\ 8i_1-18i_2=0\end{array}$                                                                                    (III)

Write the expression based on junction rule.

    $i_1=i_1-i_2+i_2$                                                                                                           (IV)

Solve equation (III) and (IV) to calculate $i_1$ and $i_2$.

    $\begin{array}{l}{i}_1=0\text{A}\\ i_2=0\text{A}\end{array}$

Thus, the current across the circuit (b) is zero.

Therefore, the power delivered to across $10\text{Ω}$ is zero in case of circuit (b).

The current flow diagram for circuit (c) is as shown in the figure below.

Figure-(3)

Write the equation of Kirchhoff’s voltage rule in the left loop.

    $\begin{array}{c}5i_1+\left(i_1-i_2\right)8-15=0\\ 5i_1+8i_1-8i_2-15=0\\ 13i_1-8i_2=0\end{array}$                                                                                               (V)

Write the equation of Kirchhoff’s voltage rule in the left loop.

    $\begin{array}{c}5i_1+\left(i_1-i_2\right)8-15=0\\ 5i_1+8i_1-8i_2-15=0\\ 13i_1-8i_2=0\end{array}$                                                                                              (VI)

Solve equation (V) and (VI) to calculate $i_1$ and $i_2$.

    $\begin{array}{l}{i}_1=2.29\text{A}\\ i_2=1.85\text{A}\end{array}$

Write the expression to obtain the power across the $10.0\Omega$ resistor.

    $P=i_2^2{R}_{10\text{Ω}}$

Here, $P$ is the power across the $10.0\text{Ω}$ resistor, $i_2$ is the current across the $10.0\text{Ω}$ resistor $R_{10\text{Ω}}$ is the $10\text{Ω}$ resistance.

Substitute $1.85\text{A}$ for $i_2$ and $10.0\text{Ω}$ for $R_{5\text{Ω}}$ in the above equation to calculate $P$.

    $\begin{array}{c}P={\left(1.85\text{A}\right)}^2\left(10.0\text{Ω}\right)\\ =\left(1.85\text{A}\right)\left(10.0\text{Ω}\right)\\ =34.33\text{W}\end{array}$

Therefore, the power delivered to the $10.0\text{Ω}$ resistor is $34.33\text{W}$ in case of circuit (c).

The current flow diagram for circuit (d) is as shown in the figure below.

Figure-(4)

Write the equation of Kirchhoff’s voltage rule in left loop.

    $\begin{array}{l}5i_1+\left(i_1-i_2\right)=0\\ 6i_1-i_2=0\end{array}$                                                                                                    (VII)

Write the equation of Kirchhoff’s voltage rule in left loop.

    $\begin{array}{l}15-10i_2-15+\left(i_1-i_2\right)8=0\\ -10i_2-8i_2+8i_1=0\\ 8i_1-18i_2=0\\ 4i_1-9i_2=0\end{array}$                                                                                (VIII)

Solve equation (VII) and (VIII) to calculate $i_1$ and $i_2$.

    $\begin{array}{l}{i}_1=0\text{A}\\ i_2=0\text{A}\end{array}$

Thus, the current across the circuit (d) is zero.

Therefore, the power delivered to across $10\text{Ω}$ is zero in case of circuit (d).

Therefore the smallest power is delivered across $10.0\text{Ω}$ resistor in case of circuit (c) which is equal to $34.33\text{W}$ as in other two cases power across the $10.0\text{Ω}$ is zero.