Question

Imagine that you have three resistors of 30 ohms

Answer 1

can obtain the following resistors by various combinations.
i) When the three resistors are connected in series
${{R}{eq}}$ = R1 + R2 + R3
= 30+ 30+ 30 = 90 Ω.
ii) If all the three resistors are connected in parallel
1/ ${{R}{eq}}$ = 1/r1 + 1/r2 + 1/r3
= 1/30 + 1/30+ 1/30
=1+1+1/30 = 3/30 = 1/10
${{R}{eq}}$ = 10 Ω.
iii) If two resistors are connected in series and the third resistor is connected in parallel to them.
${{R}{eq}}$ for series = R1 + R2
= 30+ 30 = 60 Ω.
${{R}{eq}}$ for 60 Ω, 30 Ω in parallel
1/${{R}{eq}}$ = 1/60 + 1/30 = 1+2/30 = 3/60 = 1/20
${{R}{eq}}$ = 20 Ω