Chapter 21, Problem 21CQ

To determine

(a)

The potential difference going from point $a$ to point $b$.

Answer to Problem 21CQ

  ${\text{V}}_{\text{ab}}{\text{=V}}_{\text{b}}-{\text{V}}_{\text{a}}\text{=}-{\text{I}}_1{\text{R}}_1$

Explanation of Solution

Given information:

The given figure is

Introduction:

Kirchhoff's junction rule states that the sum of all the currents entering a junction is equal to the sum of all the currents leaving the junction.

The potential difference between the two points $a$ and $b$ is calculated as

  ${\text{V}}_{\text{ab}}{\text{=V}}_{\text{b}}-{\text{V}}_{\text{a}}$

The potential difference between the two points $a$ and $b$ is calculated as

  ${\text{V}}_{\text{ab}}{\text{=V}}_{\text{b}}-{\text{V}}_{\text{a}}$

Since it is clearly seen that the current direction is from $a$ to $b$ so the potential difference between the two points $a$ and $b$ will be

  ${\text{V}}_{\text{ab}}{\text{=V}}_{\text{b}}-{\text{V}}_{\text{a}}\text{=}-{\text{I}}_1{\text{R}}_1$

Because the resistor is traversed in same direction of current so the value is negative.

Conclusion:

The potential difference going from point $a$ to point $b$ is ${\text{V}}_{\text{ab}}{\text{=V}}_{\text{b}}-{\text{V}}_{\text{a}}\text{=}-{\text{I}}_1{\text{R}}_1$.

To determine

(b)

The potential difference going from point $c$ to point $b$.

Answer to Problem 21CQ

  ${\text{V}}_{c\text{b}}{\text{=V}}_{\text{b}}-{\text{V}}_c{\text{=I}}_2{\text{R}}_3$

Explanation of Solution

Given information:

The given figure is

Introduction:

Kirchhoff's junction rule states that the sum of all the currents entering a junction is equal to the sum of all the currents leaving the junction.

The potential difference between the two points $c$ and $b$ is calculated as

  ${\text{V}}_{\text{cb}}{\text{=V}}_{\text{b}}-{\text{V}}_c$

The potential difference between the two points $c$ and $b$ is calculated as

  $V_{cb}=V_b-V_c$

Since it is clearly seen that the current direction is from $b$ to $c$ so the potential difference between the two points $c$ and $b$ will be

  ${\text{V}}_{c\text{b}}{\text{=V}}_{\text{b}}-{\text{V}}_c{\text{=I}}_2{\text{R}}_3$

Because the resistor is traversed in opposite direction of current so the value is positive.

Conclusion:

The potential difference going from point $c$ to point $b$ is $V_{cb}=V_b-V_c=I_2{R}_3$.

To determine

(c)

The potential difference going from point $e$ to point $g$

Answer to Problem 21CQ

  ${\text{V}}_{eg}{\text{=V}}_g-{\text{V}}_e{\text{=I}}_3{r}_2$

Explanation of Solution

Given information:

The given figure is

Introduction:

Kirchhoff's junction rule states that the sum of all the currents entering a junction is equal to the sum of all the currents leaving the junction.

The potential difference between the two points $e$ and $g$ is calculated as

  ${\text{V}}_{eg}{\text{=V}}_g-{\text{V}}_e$

The potential difference between the two points $a$ and $b$ is calculated as

  ${\text{V}}_{eg}{\text{=V}}_g-{\text{V}}_e$

Since it is clearly seen that the current direction is from $g$ to $e$ so the potential difference between the two points $e$ and $g$ will be

  ${\text{V}}_{eg}{\text{=V}}_g-{\text{V}}_e{\text{=I}}_3{r}_2$

Because the resistor is traversed in opposite direction of current so the value is positive.

Conclusion:

The potential difference going from point $e$ to point $g$ is ${\text{V}}_{eg}{\text{=V}}_g-{\text{V}}_e{\text{=I}}_3{r}_2$.

To determine

(d)

The potential difference going from point $e$ to point $d$

Answer to Problem 21CQ

  ${\text{V}}_{ed}{\text{=V}}_d-{\text{V}}_e\text{=}-{\text{I}}_2{r}_4$

Explanation of Solution

Given information:

The given figure is

Kirchhoff's junction rule states that the sum of all the currents entering a junction is equal to the sum of all the currents leaving the junction.

The potential difference between the two points $e$ and $d$ is calculated as

  ${\text{V}}_{ed}{\text{=V}}_d-{\text{V}}_e$

The potential difference between the two points $e$ and $d$ is calculated as

  ${\text{V}}_{\text{ab}}{\text{=V}}_{\text{b}}-{\text{V}}_{\text{a}}$

Since it is clearly seen that the current direction is from $d$ to $e$ so the potential difference between the two points $e$ and $d$ will be

  ${\text{V}}_{ed}{\text{=V}}_d-{\text{V}}_e\text{=}-{\text{I}}_2{r}_4$

Because the resistor is traversed in same direction of current so the value is negative.

Conclusion:

The potential difference going from point $e$ to point $d$ is ${\text{V}}_{ed}{\text{=V}}_d-{\text{V}}_e\text{=}-{\text{I}}_2{r}_4$.