Chapter 8.5, Problem 47PPSE

Solution

a.)

To calculate:The resistance of parallel circuit with 3 bulbs having resistance 5 ohms, 4 ohms and $2.5$ ohms using the formula $\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}$ .

The resistance of parallel circuit with 3 bulbs having resistance 5 ohms, 4 ohms and $2.5$ ohms is $\frac{20}{13}$ .

Given information:

Theresistance of parallel circuit with 3 bulbs having resistance 5 ohms, 4 ohms and $2.5$ ohms

Formula Used:

The resistance of parallel circuitwith 3 bulbs $\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}$ .

Calculation:

Consider the formula $\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}$ ,

  $\begin{array}{c}\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}=\frac{1}{\frac{R_2{R}_3+R_1{R}_3+R_1{R}_2}{R_1{R}_2{R}_3}}\\ =\frac{R_1{R}_2{R}_3}{R_2{R}_3+R_1{R}_3+R_1{R}_2}\end{array}$

Put $R_1=5$ , $R_2=4$ and $R_3=2.5$ .

  $\begin{array}{c}\frac{1}{\frac{1}{5}+\frac{1}{4}+\frac{1}{2.5}}=\frac{5\left(4\right)\left(2.5\right)}{5\left(4\right)+5\left(2.5\right)+4\left(2.5\right)}\\ =\frac{50}{20+12.5+10}\\ =\frac{50}{32.5}\\ =\frac{20}{13}\end{array}$

Thus, the resistance of parallel circuit with 3 bulbs having resistance 5 ohms, 4 ohms and $2.5$ ohms is $\frac{20}{13}$ .

b.)

To calculate: The resistance of parallel circuit with 3 bulbs if two of them have equal resistance while the resistance of third one is 3 ohms less given that resistance of parallel circuit with 3 bulbs is $1.5$ ohms using the formula $\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}$ .

The resistance of parallel circuit with 3 bulbs having resistance 5 ohms, 4 ohms and $2.5$ ohms is $\frac{20}{13}$ .

Given information:

The resistance of parallel circuit with 3 bulbs if two of them have equal resistance while the resistance of third one is 3 ohms less given that resistance of parallel circuit with 3 bulbs is $1.5$ ohms.

Formula Used:

The resistance of parallel circuit with 3 bulbs $\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}$ .

Calculation:

Consider the formula $\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}$ ,

  $\begin{array}{c}\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}=\frac{1}{\frac{R_2{R}_3+R_1{R}_3+R_1{R}_2}{R_1{R}_2{R}_3}}\\ =\frac{R_1{R}_2{R}_3}{R_2{R}_3+R_1{R}_3+R_1{R}_2}\end{array}$

Put $R_1=A$ , $R_2=A$ and $R_3=A-3$ .

  $\begin{array}{c}1.5=\frac{A\left(A\right)\left(A-3\right)}{\left(A\right)\left(A-3\right)+\left(A\right)\left(A-3\right)+A\left(A\right)}\\ =\frac{A^2\left(A-3\right)}{3A^2-6A}\\ =\frac{A^2\left(A-3\right)}{3A\left(A-2\right)}\\ =\frac{A\left(A-3\right)}{3\left(A-2\right)}\end{array}$

Multiply each side with $3\left(A-2\right)$ .

  $\begin{array}{c}1.5\left(3\right)\left(A-2\right)=\frac{A\left(A-3\right)}{3\left(A-2\right)}\left(3\left(A-2\right)\right)\\ 4.5A-9=A^2-3A\\ A^2-7.5A+9=0\\ A^2-6A-1.5A+9=0\end{array}$

  $\begin{array}{c}A\left(A-6\right)-1.5\left(A-6\right)=0\\ \left(A-6\right)\left(A-1.5\right)=0\end{array}$

Equate each value to 0.

  $\begin{array}{c}A-6=0\\ A=6\end{array}$

  $\begin{array}{c}A-1.5=0\\ A=1.5\end{array}$

Thus $A=6$ as $A$ must be greater than 3.

So,

  $\begin{array}{c}A=6\\ A=6\\ A-3=3\end{array}$

Thus, the resistance of parallel circuit with 3 bulbs if two of them have equal resistance while the resistance of third one is 3 ohms less given that resistance of parallel circuit with 3 bulbs is $1.5$ ohms is 6 ohms, 6 ohms and 3 ohms.