Chapter 21, Problem 5TP

To determine

(a)

The resistors with maximum current.

Answer to Problem 5TP

  $R4=4\Omega$ has the maximum current in the circuit.

Explanation of Solution

Given:

Parallel combination of two $20\Omega$ and one $10\Omega$ resistors is connected in series with a $4\Omega$ resistor.

Source voltage = $36\text{V}\text{.}$

Calculations:

With the given information, following circuit diagram can be drawn:

  

Here, the resistors $R\text{1=}R\text{2=20}\Omega ,R3=10\Omega \text{, and }R4=4\Omega$ .The voltage source is $36\text{V}\text{.}$

The resistor which provides the maximum current is $R4=4\Omega$ ,because the current that goes through $R4$ is the total current in the circuit and that current is equally distributed in the three branches of resistors.

Conclusion:

Thus, $R4$ is the resistor which provide the maximum current.

To determine

(b)

The resistor with maximum voltage drop.

Answer to Problem 5TP

Each of the resistor connected in parallel have the maximum voltage drop.

Explanation of Solution

Given:

Parallel combination of two $20\Omega$ and one $10\Omega$ resistors is connected in series with a $4\Omega$ resistor.

Source voltage = $36\text{V}\text{.}$

Calculations:

With the given information, following circuit diagram can be drawn:

  

Here, the resistors $R\text{1=}R\text{2=20}\Omega ,R3=10\Omega \text{, and }R4=4\Omega$ .The voltage source is $36\text{V}$.

To find the voltage drop in the above circuit, compare the voltage across of $R1,R2,R3$ with a voltage across $R4$.

So, for finding the voltage drop, find the equivalent resistance separately as follows:

  $R1,R2,\text{ and }R3$ are in parallel. Their equivalent resistance can be calculated as follows:

  $\begin{array}{l}\Rightarrow \frac{1}{R_{eq}}=\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}\text{ {Formula to find the equivalent resistance}}\\ \Rightarrow \frac{1}{R_{eq}}=\frac{1}{20}+\frac{1}{20}+\frac{1}{10}\text{ {where,}R\text{1=}R\text{2=20}\Omega ,R3=10\Omega \}\\ \Rightarrow \frac{1}{R_{eq}}=\frac{1+1+2}{20}\\ \Rightarrow \frac{1}{R_{eq}}=\frac{1+1+2}{20}\\ \Rightarrow \frac{1}{R_{eq}}=\frac{4}{20}\\ \Rightarrow R_{eq}=5\Omega \end{array}$

Conclusion:

Thus, the equivalent resistance of $5\Omega$ is greater than the $R4.$ So, each of the resistor connected in parallel will have the maximum voltage drop.

To determine

(c)

The power dissipated in each of the resistor and total power dissipated in all resistor.

Answer to Problem 5TP

20 W in each $20\Omega$ resistor, 40 W in $10\Omega$ resistor. Total = 144 W.

Output power source = 144 W.

Yes, they are equal as per law of conservation of energy.

Explanation of Solution

Given:

Parallel combination of two $20\Omega$ and one $10\Omega$ resistors is connected in series with a $4\Omega$ resistor.

Source voltage = $36\text{V}\text{.}$

Calculations:

With the given information, following circuit diagram can be drawn:

  

Here, the resistors $R\text{1=}R\text{2=20}\Omega ,R3=10\Omega \text{, and }R4=4\Omega$.

The voltage source is $36\text{V}\text{.}$

To find the power dissipated in each of the resistor and the total power dissipated in each of the resistor, follow some steps one by one as below:

1. Find the equivalent resistance and then find the total current.

  $\begin{array}{l}\Rightarrow \frac{1}{R}=\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}\text{ {Formula to find the equivalent resistance}}\\ \Rightarrow \frac{1}{R}=\frac{1}{20}+\frac{1}{20}+\frac{1}{10}\text{ {where,}R\text{1=}R\text{2=20}\Omega ,R3=10\Omega \}\\ \Rightarrow \frac{1}{R}=\frac{1+1+2}{20}\\ \Rightarrow \frac{1}{R}=\frac{1+1+2}{20}\\ \Rightarrow \frac{1}{R}=\frac{4}{20}\\ \Rightarrow R=5\Omega \\ \text{Parallel combination of the circuit with series resistance}\text{.}\\ R_{eq}=R+R4=5\Omega +4\Omega =9\Omega \text{ {}\therefore \text{,}R4=4\Omega \text{}}\\ \Rightarrow \text{Total current:}\\ I=\frac{V}{R}=\frac{36}{9}=4A\text{ {Voltage source(}V\text{)=36 ,}R=9\}\end{array}$

  $\text{Step 2: Voltage across parallel combination of resistance (}R1,R2,R3\text{) are:}$

2. Voltage across parallel combination of resistance.

  $\begin{array}{l}\Rightarrow V=I\times R\\ \Rightarrow V=4\times 5\text{ {where,}I=4\text{ and }R=5\text{}}\\ \Rightarrow V=20\text{ V}\end{array}$

3. Calculate the power dissipated in each resistor are:

  $\begin{array}{l}\Rightarrow P_1=\frac{V^2}{R}=\frac{{(20)}^2}{20}=\frac{400}{20}=20\text{ watt}\\ \Rightarrow P_2=\frac{V^2}{R}=\frac{{(20)}^2}{20}=\frac{400}{20}=20\text{ watt}\\ \Rightarrow P_3=\frac{V^2}{R}=\frac{{(20)}^2}{10}=\frac{400}{10}=40\text{ watt}\\ \text{Power dissipated across the resistance (}R4\text{) are:}\\ $ P_4=I^{2}R={(4)}^2\cdot 4=64\text{ watt}$\end{array}$

4. Total power dissipated in all the resistor.

  $\begin{array}{l}\Rightarrow P=P_1+P_2+P_3+P_4\\ \Rightarrow P=20+20+40+64\\ \Rightarrow P=144\text{ watt}\end{array}$

5. Find the total power output of the source as follows:

  $\begin{array}{l}\Rightarrow \text{Total power output source(}P)=IV\text{ {Substitute the value of }I=4A\text{ and }V=36V\text{}}\\ \Rightarrow \text{ =4}\times \text{36}\\ \Rightarrow \text{ =144 W}\end{array}$

Conclusion:

The power dissipated of all the resistor and the total power output source are equal by the law of conservation of energy.

To determine

(e)

The values of all the resistors and the source voltage are doubled.

Answer to Problem 5TP

No effect.

Explanation of Solution

Given:

Parallel combination of two $20\Omega$ and one $10\Omega$ resistors is connected in series with a $4\Omega$ resistor.

Source voltage = $36\text{V}$.

Calculations:

With the given information, following circuit diagram can be drawn:

  

Here, the resistors $R\text{1=}R\text{2=20}\Omega ,R3=10\Omega \text{, and }R4=4\Omega$ .The voltage source is $36\text{V}$.

If the values of all the resistors and the source voltage are doubled, then there would be no effect on the current.