Chapter 19, Problem 6P

(a)

To determine

The total resistance of the circuit when three bulbs of $45\Omega$ and three bulbs of $75\Omega$ are connected in series.

Answer to Problem 6P

The total resistance of the circuit in series connection is $360\Omega$ .

Explanation of Solution

Given information:

Three lightbulb has resistance of $45\Omega$ each and three lightbulbs has resistance of $75\Omega$ each.

Formula used:

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

  $R_{\text{S}}=R+R+R+R+R+R+R$............. (1)

Where,

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

Calculation:

Substitute the value in equation (1).

  $\begin{array}{c}{R}_{\text{S}}=45+45+45+75+75+75\\ =135+225\\ =360\Omega \end{array}$

Therefore, the total resistance of the circuit in series connection is $360\Omega$ .

(b)

To determine

The total resistance of the circuit when three bulbs of $45\Omega$ and three bulbs of $75\Omega$ are connected in parallel to each other.

Answer to Problem 6P

The total resistance of the circuit in parallel connection is $9.37\Omega$ .

Explanation of Solution

Given information:

Three lightbulb has resistance of $45\Omega$ each and three lightbulbs has resistance of $75\Omega$ each.

Formula used:

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

  $\frac{1}{R_{\text{P}}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{R}$............. (2)

Where,

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

Calculation:

Substitute the value in equation (2).

  $\begin{array}{c}\frac{1}{R_{\text{P}}}=\frac{1}{45}+\frac{1}{45}+\frac{1}{45}+\frac{1}{75}+\frac{1}{75}+\frac{1}{75}\\ \frac{1}{R_{\text{P}}}=\frac{3}{45}+\frac{3}{75}\\ R_{\text{P}}=\frac{45\times 75}{3\left(75+45\right)}\\ =9.37\Omega \end{array}$

Therefore, the total resistance of the circuit in parallel connection is $9.37\Omega$ .