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

Step-by-step explanation:

The resistances of the given resistors are:

`\begin{gathered} R_1=7.50*10^2Ω \\ R_1=0.75*10^3=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} (1)/(R_(eq))=(1)/(R_1)+(1)/(R_2)+(1)/(R_3) \\ \\ (1)/(R_(eq))=(1)/(0.75)+(1)/(2.4)+(1)/(4.5) \\ \\ (1)/(R_(eq))=1.972 \\ \\ R_(eq)=(1)/(1.972) \\ \\ R_(eq)=0.51kΩ \end{gathered}`