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

Step-by-step explanation

the formula to add resistances in parallel is

`(1)/(R_(eq))=(1)/(R_1)+(1)/(R_2)+....(1)/(R_n)`

so

Step 1

a) let

b) now, replace in the formula and calculate

`\begin{gathered} (1)/(R_(eq))=(1)/(R_(1))+(1)/(R_(2))+(1)/(R_3) \\ (1)/(R_(eq))=\frac{1}{0.75\text{ k}\Omega}+(1)/(2.4k\Omega)+(1)/(4.5k\Omega) \\ (1)/(R_(eq))=1.972\text{ k}\Omega \\ so \\ R_(eq)=(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