Question

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

Answer 1

507 ohms

Explanation

the formula to add resistances in parallel is

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+....\frac{1}{R_n}$

so

Step 1

a) let

$\begin{gathered} R_1=7.5*10^2\text{ \Omega=0.75*10}^3=0.75\text{ K\Omega} \\ R_2=2.4\text{ K\Omega} \\ R_3=4.5\text{ K\Omega} \end{gathered}$

b) now, replace in the formula and calculate

$\begin{gathered} \frac{1}{R_{eq}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_3} \\ \frac{1}{R_{eq}}=\frac{1}{0.75\text{ k}\Omega}+\frac{1}{2.4k\Omega}+\frac{1}{4.5k\Omega} \\ \frac{1}{R_{eq}}=1.972\text{ k}\Omega \\ so \\ R_{eq}=\frac{1}{1.972}k\Omega \\ R_{eq}=0.507\text{ k}\Omega \\ \end{gathered}$

finally to convert from kiloohms into ohms, multiply by 1000, so

so

$\begin{gathered} Re_q=0.507\text{ k}\Omega\text{ =\lparen0.507*1000\rparen}\Omega \\ R_{eq}=507.\Omega \end{gathered}$

so

the answer is 507 ohms