Chapter 19, Problem 9P

(a)

To determine

The maximum resistance.

Answer to Problem 9P

The maximum resistance is $2820\Omega$ .

Explanation of Solution

Given information:

Resistance of the first register, $R_1=680\Omega$

Resistance of the second register, $R_2=940\Omega$

Resistance of the third register, $R_3=1.20\text{k}\Omega$

Formula used:

Write the expression to calculate the total resistance of the circuit in series connection.

  $R_{\text{S}}=R_1+R_2+R_3$............ (1)

Where,

• The total resistance of the circuit is $R_{\text{S}}$ in series connection.

Calculation:

The maximum resistance can be obtained when the resistors are connected in series.

Substitute the value in equation (1).

  $\begin{array}{c}{R}_{\text{S}}=680+940+1.20\times {10}^3\\ =2820\Omega \end{array}$

Therefore, the maximum resistance is $2820\Omega$ .

(b)

To determine

The minimum resistance.

Answer to Problem 9P

The minimum resistance is $296.93\Omega$ .

Explanation of Solution

Given information:

Resistance of the first register, $R_1=680\Omega$

Resistance of the second register, $R_2=940\Omega$

Resistance of the third register, $R_3=1.20\text{k}\Omega$

Formula used:

Write the expression to calculate the total resistance of the circuit in parallel connection.

  $\frac{1}{R_{\text{P}}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$............ (2)

Where,

• The total resistance of the circuit is $R_{\text{P}}$ in parallel connection.

Calculation:

The minimum resistance can be obtained when the resistors are connected in parallel.

Substitute the value in equation (2).

  $\begin{array}{c}\frac{1}{R_{\text{P}}}=\frac{1}{680}+\frac{1}{940}+\frac{1}{1.20\times {10}^3}\\ R_{\text{P}}=296.93\Omega \end{array}$

Therefore, the minimum resistance is $296.93\Omega$ .