Question

How can three resistors R₁= 3 Ω , R2= 3 Ω , and R3= 5 Ω be connected to give a total resistance of 6.5 Ω?
1. When R1 and R2 are connected in parallel with R3 in series
2. When R1 and R3 are connected in parallel with R2 in series
3. When R2 and R3 are connected in parallel with R1 in series
4. All connected in series

Answer 1

Correct Answer - Option 1 : When R1 and R2 are connected in parallel with R3 in series

R₁ = 3 ohm           

R₂ = 3 ohm           

R₃ = 5 ohm           

When R1 and R2 are connected in parallel with R3 in series we get                 

1/R = 1/R1 + 1/R2                 

= 1/3 + 1/3                 

G = 1.5            

Thus, R = 1.5 Ohm        

Resistance in series = R + R₃            

= 1.5 + 5                

= 6.5 ohm