Chapter 25, Problem 90P

(a)

To determine

The current in each resistor

Answer to Problem 90P

The current in each resistor is $I_{1,2\Omega }=2\text{A, }{I}_{2\Omega }=-1\text{A}\text{and}{I}_{6\Omega }=1\text{A}$ .

Explanation of Solution

Given:

The voltage of the battery is $\epsilon =8\text{V}$ .

The circuit is shown in figure.

  

Figure (1)

Formula used:

The expression for the Kirchhoff’s junction ruleis given by,

  $I_{in}=I_{out}$

The expression for the Kirchhoff’s loop rule is given by:

  ${\displaystyle \sum V}=0$

Calculation:

Applying the Kirchhoff’s loop rule to the outside loop of the circuit:

  $\begin{array}{l}\text{8}\text{V}+\text{4}\text{V}-\left(\text{1}{\text{Ω×I}}_{\text{1,2}\text{Ω}}\right)-\left(\text{6}{\text{Ω×I}}_{\text{6}\text{Ω}}\right)-\left(\text{2}{\text{Ω×I}}_{\text{1,2}\text{Ω}}\right)=\text{0}\\ 3\Omega \times I_{1,2\Omega }+6\Omega \times I_{6\Omega }=12V\end{array}$ ..... (1)

Similarly, apply the Kirchhoff’s loop rule to the inside loop at the LHS of the circuit:

  $\begin{array}{l}\text{8}\text{V}+\text{4}\text{V}-\left(\text{1}{\text{Ω×I}}_{\text{1,2}\text{Ω}}\right)-\left(2{\text{Ω×I}}_{\text{2}\text{Ω}}\right)-\left(\text{2}{\text{Ω×I}}_{\text{1,2}\text{Ω}}\right)-4\text{V}=0\\ 3\Omega \times I_{1,2\Omega }+2\Omega \times I_{2\Omega }=8\text{V}\end{array}$

Apply the Kirchhoff’s junction rule,

  $\begin{array}{l}{I}_{6\Omega }=I_{2\Omega }+I_{1,2\Omega }\\ 3\Omega \times I_{1,2\Omega }+6\Omega \times \left(I_{2\Omega }+I_{1,2\Omega }\right)=12V\end{array}$ (2)

On solving (1) and (2) equation,

  $\begin{array}{c}{I}_{1,2\Omega }=2\text{A}\\ {\text{I}}_{2\Omega }=-1\text{A}\\ I_{6\Omega }=2\text{A}+\left(-1\text{A}\right)\\ =1\text{A}\end{array}$

Conclusion:

Therefore, the current in each resistor is $I_{1,2\Omega }=2\text{A, }{I}_{2\Omega }=-1\text{A}\text{and}{I}_{6\Omega }=1\text{A}$ .

(b)

To determine

The power supplied by each source of emf.

Answer to Problem 90P

The power supplied by each source of emf is $P_{8V}=16\text{W}$ and $P_{4V}=-4\text{W}$ .

Explanation of Solution

Formula used:

The expression for the power supplied by emf source is given by,

  $P=IV$

Calculation:

The power delivered by the source of emf $\text{8}\text{V}$ is calculated as,

  $\begin{array}{c}P=IV\\ P_{8V}=I_{1,2\Omega }\times 8\text{V}\\ \text{=}\left(2\text{A}\right)\times \left(8\text{V}\right)\\ =16\text{W}\end{array}$

The power delivered by source of 4V is calculated as,

  $\begin{array}{c}{P}_{4V}=I_{2\Omega }\times 4\text{V}\\ \text{=}\left(-1\text{A}\right)\times \left(4\text{V}\right)\\ =-4\text{W}\end{array}$

Conclusion:

Therefore, the power supplied by each source of emf is $P_{8V}=16\text{W}$ and $P_{4V}=-4\text{W}$ .

(c)

To determine

The power delivered to each resistor

Answer to Problem 90P

The power delivered to each resistor is $P_{1\Omega }=2\text{W}$ , $P_{2\Omega }=4\text{W}$ , $P_{2\Omega }=-2\text{W}$ and $P_{6\Omega }=6\text{W}$ .

Explanation of Solution

Formula used:

The expression for the power delivered by the resistoris given by,

  $P=I{R}^2$

Calculation:

The delivered by the $1\Omega$ resistor is given by

  $\begin{array}{c}{P}_{1\Omega }=I_{1,2\Omega }{\left(R_{1\Omega }\right)}^2\\ =\left(2\text{A}\right)\left(1\Omega \right)\\ =2\text{W}\end{array}$

Similarly,

  $\begin{array}{c}{P}_{2\Omega }=I_{1,2\Omega }{\left(R_{2\Omega }\right)}^2\\ =\left(2\text{A}\right)\left(2\Omega \right)\\ =4\text{W}\end{array}$

  $\begin{array}{c}{P}_{2\Omega }=I_{2\Omega }{\left(R_{2\Omega }\right)}^2\\ =\left(-1\text{A}\right)\left(2\Omega \right)\\ =-2\text{W}\end{array}$

  $\begin{array}{c}{P}_{6\Omega }=I_{6\Omega }{\left(R_{6\Omega }\right)}^2\\ =\left(1\text{A}\right)\left(6\Omega \right)\\ =6\text{W}\end{array}$

Conclusion:

Therefore, the power delivered to each resistor is $P_{1\Omega }=2\text{W}$ , $P_{2\Omega }=4\text{W}$ , $P_{2\Omega }=-2\text{W}$ and $P_{6\Omega }=6\text{W}$ .