Question

(a) What is the resistance (in kΩ) of a 7.50 ✕ 102 Ω, a 2.40 kΩ, and 4.50 kΩ resistor connected in series? kΩ(b) What is the resistance (in Ω) of a 7.50 ✕ 102 Ω, a 2.40 kΩ, and 4.50 kΩ resistor connected in parallel? Ω†

Answer 1

Answer:

a) The equivalent resistance in series = 7.65 kΩ

b) The equivalent resistance in parallel = 0.51 kΩ

Explanation:

The resistances of the given resistors are:

$\begin{gathered} R_1=7.50\times10^2Ω \\ R_1=0.75\times10^3=0.75kΩ \end{gathered}$$\begin{gathered} R_2=2.40kΩ \\ R_3=4.50kΩ \end{gathered}$

The equivalent resistance in series:

$\begin{gathered} R_{eq}=R_1+R_2+R_3 \\ \\ R_{eq}=0.75kΩ+2.40kΩ+4.50kΩ \\ \\ R_{eq}=7.65kΩ \end{gathered}$

b) The equivalent resistance in parallel

$\begin{gathered} \frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3} \\ \\ \frac{1}{R_{eq}}=\frac{1}{0.75}+\frac{1}{2.4}+\frac{1}{4.5} \\ \\ \frac{1}{R_{eq}}=1.972 \\ \\ R_{eq}=\frac{1}{1.972} \\ \\ R_{eq}=0.51kΩ \end{gathered}$