Question

Three resistance of 12,15,20 ohms are connected first in series and then in parallel . What is the combined resistance in each case

Answer 1

Total resistance in series
R= R1+R2+R3
R=12+15+20
R=47ohm


Total resistance in parallel
1/R= 1/R1+1/R2+1/R2
1/R=1/12+1/15+1/20
1/R= 5+4+3/60
= 12/60

= 5ohm

ANSWER

Answer 2

$\huge{SOLUTION:-}$

$\red{SERIES}$

The effective resistance of a circuit when the resistances are connected in series is given by,

$\boxed{R_{s}=R_{1}+R_{2}+R_{3}}$

So effective Resistance is

12+15+20 = 47 ohms

$\red{PARALLEL}$

The effective resistance of a circuit when the resistances are connected in parallel is given by:-

$\boxed{\frac{1}{R_{p}}} = {\frac{1}{R_{1}}} + {\frac{1}{R_{2}}}$

Hence, effective resistance is

$\frac{1}{12}$ + $\frac{1}{15}$ + $\frac{1}{20}$

= $\frac{5+4+3}{60}$

= $\frac{12}{60}$

Hence resistance is

$\frac{60}{12}$ = 5ohms