A 35 kΩ is connected in series with a resistance of 40 kΩ. What resistance R must be connected in parallel with the combination so that the equivalent resis...

Answer Details

For the combination in series;

⇒R1 = 35kΩ + 40kΩ = 75kΩ

R is combined with 75kΩ in parallel to give 25kΩ

=  1Req $\frac{\mathrm{<br>}}{}$ = 1R $\frac{\mathrm{<br>}}{}$ + 1R $\frac{\mathrm{<br>}}{}$ 

=   125 $\frac{\mathrm{<br>}}{}$     = 1R $\frac{\mathrm{<br>}}{}$ + 175 $\frac{\mathrm{<br>}}{}$ 

=  125 $\frac{\mathrm{<br>}}{}$     - 175 $\frac{\mathrm{<br>}}{}$ + 1R $\frac{\mathrm{<br>}}{}$ 

=   3−175 $\frac{\mathrm{<br>}}{}$ = 1R $\frac{\mathrm{<br>}}{}$ 

=   275 $\frac{\mathrm{<br>}}{}$   = 1R $\frac{\mathrm{<br>}}{}$ 

=   752 $\frac{\mathrm{<br>}}{}$  = R

;      R = 37.5k Ω