Chapter 26, Problem 42SP

For the circuit of Fig. 26-6, find the potential difference from (a)A to B, (b)B to C, and (c)C to A. Notice that the current is given as 2.0 A.

To determine

The potential difference from $A$ to $B$ for the circuit shown in Fig. 26-6 if the current flowing in the circuit is $2.0\text{ A}$.

Answer to Problem 42SP

Solution:

$-48\text{ V}$

Explanation of Solution

Given data:

The given circuit diagram is

Formula used:

The expression for the potential difference across the resistance from Ohm’s law is written as

$V=IR$

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

The Kirchhoff’s voltage law states that in closed mesh, the sum of the voltage drop across the resistors is equal to the sum of emf of the source.

Mathematically it is represented as

${\displaystyle \sum IR+{\displaystyle \sum \epsilon }}=0$

Here, $\epsilon$ is electromotive force available in the loop.

Explanation:

Draw the circuit diagram to analyse the problem:

Here, the direction of the loop is opposite to the direction of the current. Put plus and minus signs on resistors according to the direction of current, the positive sign at the entering the current and negative sign where current leaves.

Apply the Kirchhoff’s voltage rule to calculate the value of potential from $B$ to $A$:

$-I\left(8.0\Omega \right)-12\text{ V}-I\left(9.0\Omega \right)+6.0\text{ V}-I\left(4.0\Omega \right)+V_{BA}=0$

Here, $V_{BA}$ is potential from $B$ to $A$.

Rearrange the expression for $V_{BA}$:

$V_{BA}=I\left(8.0\Omega \right)+12\text{ V}+I\left(9.0\Omega \right)-6.0\text{ V}+I\left(4.0\Omega \right)$

Substitute $2.0\text{ A}$ for $I$

$\begin{array}{c}{V}_{BA}=\left(2.0\text{ A}\right)\left(8.0\Omega \right)+12\text{ V}+\left(2.0\text{ A}\right)\left(9.0\Omega \right)-6.0\text{ V}+\left(2.0\text{ A}\right)\left(4.0\Omega \right)\\ =48\text{ V}\end{array}$

Write the expression for the potential from $A$ to $B$:

$V_{AB}=-V_{BA}$

Substitute $48\text{ V}$ for $V_{BA}$

$V_{AB}=-48\text{ V}$

Understand that the direction of the loop is $B$ to $A$. So, $V_{AB}=-V_{BA}$.

Conclusion:

Hence, the potential difference from $A$ to $B$ for the circuit is $-48\text{ V}$.

To determine

The potential difference from $B$ to $C$ for the circuit shown in Fig. 26-6 if the current flowing in the circuit is $2.0\text{ A}$.

Answer to Problem 42SP

Solution:

$\text{+28 V}$

Explanation of Solution

Given data:

The given circuit diagram is

Formula used:

The expression for the potential difference across the resistance from Ohm’s law is written as

$V=IR$

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

The Kirchhoff’s voltage law states that in closed mesh, the sum of the voltage drop across the resistors is equal to the sum of emf of the source.

Mathematically it is represented as

${\displaystyle \sum IR+{\displaystyle \sum \epsilon }}=0$

Here, $\epsilon$ is electromotive force available in the loop.

Explanation:

Redraw the circuit diagram to analyse the problem:

Apply the Kirchhoff’s voltage rule to calculate the value of potential from $B$ to $C$:

$-I\left(8.0\Omega \right)-12\text{ V}+V_{BC}=0$

Here, $V_{BC}$ is potential from $B$ to $C$.

Rearrange the expression for $V_{BC}$

$V_{BC}=I\left(8.0\Omega \right)+12\text{ V}$

Substitute $2.0\text{ A}$ for $I$

$\begin{array}{c}{V}_{BC}=\left(2.0\text{ A}\right)\left(8.0\Omega \right)+12\text{ V}\\ \text{=28 V}\end{array}$

Conclusion:

Hence, the potential difference from $B$ to $C$ for the circuit is $\text{+28 V}$.

To determine

The potential difference from $C$ to $A$ for the circuit shown in Fig. 26-6 if the current flowing in the circuit is $2.0\text{ A}$.

Answer to Problem 42SP

Solution:

$\text{+20 V}$

Explanation of Solution

Given data:

The given circuit diagram is

Formula used:

The expression for the potential difference across the resistance from Ohm’s law is written as

$V=IR$

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

The Kirchhoff’s voltage law states that in closed mesh, the sum of the voltage drop across the resistors is equal to the sum of emf of the source.

Mathematically it is represented as

${\displaystyle \sum IR+{\displaystyle \sum \epsilon }}=0$

Here, $\epsilon$ is electromotive force available in the loop.

Explanation:

Redraw the circuit diagram to analyse the problem:

Apply the Kirchhoff’s voltage rule to calculate the value of potential from $C$ to $A$:

$-I\left(9.0\Omega \right)+6\text{ V}-I\left(4.0\Omega \right)+V_{CA}=0$

Here, $V_{CA}$ is potential from $C$ to $A$.

Rearrange the expression for $V_{CA}$

$V_{CA}=I\left(9.0\Omega \right)-6\text{ V}+I\left(4.0\Omega \right)$

Substitute $2.0\text{ A}$ for $I$

$\begin{array}{c}{V}_{CA}=\left(2.0\text{ A}\right)\left(9.0\Omega \right)-6\text{ V}+I\left(4.0\Omega \right)\\ =+20\text{ V}\end{array}$

Conclusion:

Hence, the potential difference from $C$ to $A$ for the circuit is $\text{+20 V}$.