Question

Problem shown on attached image.
70 points!
Need it by November 21st, 10 am EST!

Answer 1

`\huge\boxed{\sf R_T=(3+i)/(16-2i)}`

Explanation:

Given data:

`R_1=4+6i \Omega\\\\R_2=2-4i \Omega`

Required:

`R_T=?`

Formula:

`\displaystyle R_T=(1)/(R_1) +(1)/(R_2)`

Solution:

Put the given data in above formula

`\displaystyle R_T=(1)/(4+6i) +(1)/(2-4i) \\\\Taking \ LCM\\\\R_T=(2-4i+4+6i)/((4+6i)(2-4i)) \\\\R_T=(2+4-4i+6i)/(8-16i+12i-24i^2) \\\\\because i^2=-1\\\\ R_T=(6+2i)/(8-4i-24(-1)) \\\\R_T=(6+2i)/(8+24-4i) \\\\R_T=(6+2i)/(32-4i) \\\\R_T=(2(3+i))/(2(16-2i)) \\\\R_T=(3+i)/(16-2i) \\\\\rule[225]{225}{2}`