Question

How can three resistors of resistances 2 Ω , 3 Ω , and 6 Ω be connected to give a total resistance of (a) 4 Ω , (b) 1 Ω?

Answer 1

R1 = 2 ohm          

R2 = 3 ohm          

R3 = 6 ohm          

(a)  When R2 and R3 are connected in parallel with R1 in series we get                              1/R = 1/R2 + 1/R3                

g= 1/3 + 1/6                

= 1/2                

Thus, R = 2 ohm                

Resistance in series = R + R1                

= 2 + 2                

= 4 ohm

          (b)  When R1,R2, R3 are connected in parallel we get                

1/R = 1/R1 + 1/R2 + 1/R3                

= 1/2 + 1/3 + 1/6                

= 1 ohm.