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Q21CQ

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Found in: Page 772

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# (a) What is the potential difference going from point $$a$$ to point $$b$$ in Figure $$21.47$$? (b) What is the potential difference going from $$c$$to $$b$$? (c) From $$e$$ to$$g$$? (d) From $$e$$ to $$d$$?

(a) From point $$a$$ to $$b$$ the potential difference is obtained as: $${V_{ab}}{\rm{ }} = {\rm{ }} - 7.64{\rm{ }}V$$.

(b) From point $$c$$ to $$b$$ the potential differenceis obtained as: $${V_{cb}}{\rm{ }} = {\rm{ }} - 6.48{\rm{ }}V$$.

(c) From point $$e$$ to $$g$$ the potential differenceis obtained as: $${V_{eg}}{\rm{ }} = {\rm{ }} - 2.7{\rm{ }}V$$.

(d) From point $$e$$ to $$d$$ the potential differenceis obtained as: $${V_{ed}}{\rm{ }} = {\rm{ }}23.39{\rm{ }}V$$.

See the step by step solution

## Step 1: Define Circuit

An electronic circuit is made up of individual electronic components including resistors, transistors, capacitors, inductors, and diodes that are linked by conductive wires or traces.

## Step 2: Concept

Considering an electric circuit which has two loops and then calculating the potential differences at various points. To calculate the potential drops, we are suppose to calculate three currents which goes through the circuit. To do this, we first do the simplification of the circuit by adding all resistances together on the path of one current. We define the resistances as:

\begin{aligned}{}R{\rm{ }} & = {\rm{ }}{R_1} + {r_1} + {R_4}\\ & = {\rm{ }}(20 + 0.5 + 15){\rm{ }}\Omega \\ & = {\rm{ }}35.5{\rm{ }}\Omega \end{aligned}

\begin{aligned}{}{R'}{\rm{ }} & = {\rm{ }}{R_2} + {r_2}\\ & = {\rm{ }}(6 + 0.25){\rm{ }}\Omega \\ & = {\rm{ }}6.25{\rm{ }}\Omega \end{aligned}

\begin{aligned}{}{R^{''}}{\rm{ }} & = {\rm{ }}{R_3} + {r_3} + {r_4}\\ & = {\rm{ }}(8 + 0.5 + 0.75){\rm{ }}\Omega \\ & = {\rm{ }}9.25{\rm{ }}\Omega \end{aligned}

To obtain the currents, we use the Kirchhoff's rules. We then can use the junction rule for the junction $${\rm{b}}$$, which then gives:

$${I_1}{\rm{ }} = {\rm{ }}{I_2} + {I_3}$$

The junction $${\rm{e}}$$ must also have given us the same condition. The two other conditions (we need three for three currents) are the loop conditions for the loops $${\rm{abef}}$$and $${\rm{bcde}}$$. We can also use the loop rule for the loop $${\rm{abcdef}}$$, but it will not give us any new information. We therefore have:

$${E_1} - {I_1}R - {I_3}{R'} + {E_2}{\rm{ }} = {\rm{ }}0$$

$$- {I_2}{R^{''}} + {E_3} - {E_4} + {I_3}{R'} - {E_2}{\rm{ }} = {\rm{ }}0$$

The voltage across a resistor drops when it is moved in the direction of current. If it goes in the opposite direction, the voltage difference will be $$+ IR$$. When it is gone from the positive to the negative pole of the battery , the voltage difference will be negative.

## Step 3: Obtaining the values

Inserting the first condition into the second to get the third condition.

Then we obtain:

\begin{aligned}{}{E_1} - \left( {{I_2} + {I_3}} \right)R - {I_3}{R'} + {E_2}{\rm{ }} & = {\rm{ }}0\\ - {I_2}{R{''}} + {E_3} - {E_4} + {I_3}{R'} - {E_2}{\rm{ }} & = {\rm{ }}0\\{I_2}R + {I_3}\left( {R + {R'}} \right){\rm{ }} & = {\rm{ }}{E_1} + {E_2}\\{I_2}{R^{''}} - {I_3}{R'}{\rm{ }} & = {\rm{ }} - {E_2} + {E_3} - {E_4}\\{I_2}{\rm{ }} & = {\rm{ }} - {I_3}\frac{{R + {R'}}}{R} + \frac{{{E_1} + {E_2}}}{R}\\{I_2}{\rm{ }} & = {\rm{ }}{I_3}\frac{{{R'}}}{{{R{''}}}} - \frac{{{E_2} - {E_3} + {E_4}}}{{{R^{''}}}}\\ - {I_3}\frac{{R + {R'}}}{R} + \frac{{{E_1} + {E_2}}}{R}{\rm{ }} & = {\rm{ }} + {I_3}\frac{{{R'}}}{{{R{''}}}} - \frac{{{E_2} - {E_3} + {E_4}}}{{{R{''}}}}\\{I_3}\left( {\frac{{{R'}}}{{{R{''}}}} + \frac{{R + {R'}}}{R}} \right){\rm{ }} & = {\rm{ }}\frac{{{E_1} + {E_2}}}{R} + \frac{{{E_2} - {E_3} + {E_4}}}{{{R{''}}}}\end{aligned}

The currents are no being evaluated:

\begin{aligned}{}{I_3}{\rm{ }} & = {\rm{ }}\frac{{\left( {{E_1} + {E_2}} \right){R{''}} + \left( {{E_2} - {E_3} + {E_4}} \right)R}}{{R{R'} + R{R{''}} + {R'}{R{''}}}}\\ & = {\rm{ }}\frac{{((18 + 3)V) \times (9.25{\rm{ }}\Omega ) + ((3 - 12 + 24)V) \times (35.5{\rm{ }}\Omega )}}{{(35.5{\rm{ }}\Omega ) \times (6.25{\rm{ }}\Omega ) + (35.5{\rm{ }}\Omega ) \times (9.25{\rm{ }}\Omega ) + (6.25{\rm{ }}\Omega ) \times (9.25{\rm{ }}\Omega )}}\\ & = {\rm{ }}1.195\;{\rm{ }}A\end{aligned}

\begin{aligned}{}{I_2} & = {\rm{ }}{I_3}\frac{{{R'}}}{{{R{''}}}}{\rm{ }} - {\rm{ }}\frac{{{E_2} - {E_3} + {E_4}}}{{{R^{''}}}}\\ & = {\rm{ }}1.195{\rm{ }} \times {\rm{ }}\frac{{6.25{\rm{ }}\Omega }}{{9.25{\rm{ }}\Omega }} - \frac{{(3{\rm{ }} - {\rm{ }}12{\rm{ }} + {\rm{ }}24)V}}{{9.25{\rm{ }}\Omega }}\\ & = {\rm{ }} - 0.81\;A\end{aligned}

\begin{aligned}{}{I_1}{\rm{ }} & = {\rm{ }}{I_1}{\rm{ }} + {\rm{ }}{I_2}\\ & = {\rm{ }}(1.195{\rm{ }} - {\rm{ }}0.81)A\\ & = {\rm{ }}0.382\;A\end{aligned}

## Step 4: Evaluating potential difference from point $$a$$ to $$b$$

The fact that the sign of $${I_2}$$ is a negative one then it means that we assumed the wrong direction of this current.

1. The voltage drop going from point $$a$$ to $$b$$ is:

\begin{aligned}{}{V_{ab}}{\rm{ }} & = {\rm{ }} - {I_1}{R_1}\\ & = {\rm{ }} - (0.382{\rm{ }}A) \times (20{\rm{ }}\Omega )\\ & = {\rm{ }} - 7.64{\rm{ }}V\end{aligned}

## Step 5: Evaluating potential difference from point $$c$$ to $$b$$

b. The voltage drop going from point$$c$$ to $$b$$is:

\begin{aligned}{}{V_{cb}}{\rm{ }} & = {\rm{ }}{I_2}{R_3}\\& = {\rm{ }}( - 0.81{\rm{ }}A) \times (8{\rm{ }}\Omega )\\ & = {\rm{ }} - 6.48{\rm{ }}V\end{aligned}

## Step 6: Evaluating potential difference from point $$e$$ to $$g$$

c. The voltage drop going from point$$e$$ to $$g$$ is:

\begin{aligned}{}{V_{eg}}{\rm{ }} & = {\rm{ }}{I_3}{r_2}{\rm{ }} - {\rm{ }}{E_3}\\ & = {\rm{ }}(1.95{\rm{ }}A) \times (0.25{\rm{ }}\Omega ){\rm{ }} - 3{\rm{ }}V\\ & = {\rm{ }} - 2.7{\rm{ }}V\end{aligned}

## Step 7: Evaluating potential difference from point $$e$$ to $$d$$

d. The voltage drop going from point$$e$$ to $$d$$ is:

\begin{aligned}{}{V_{ed}}{\rm{ }} & = {\rm{ }}{I_2}{r_4}{\rm{ }} + {\rm{ }}{E_4}\\& = {\rm{ }}( - 0.81{\rm{ }}A) \times (0.75{\rm{ }}\Omega ){\rm{ }} + 24\\ & = {\rm{ }}23.39{\rm{ }}V\end{aligned}

Therefore, we get:

(a) Potential difference from point $$a$$ to $$b$$ is: $${V_{ab}}{\rm{ }} = {\rm{ }} - 7.64{\rm{ }}V$$.

(b) Potential difference from point$$c$$ to $$b$$ is: $${V_{cb}}{\rm{ }} = {\rm{ }} - 6.48{\rm{ }}V$$.

(c) Potential difference from point$$e$$ to $$g$$ is: $${V_{eg}}{\rm{ }} = {\rm{ }} - 2.7{\rm{ }}V$$.

(d) Potential difference from point$$e$$ to $$d$$ is: $${V_{ed}}{\rm{ }} = {\rm{ }}23.39{\rm{ }}V$$.