Question

You are given three resistors each of 3Ω and you are asked to get all possible values of resistance when you connect them in different combinations. How many values of resistance can you get?

Answer 1

Here: R₁ = 3Ω 

R₂ = 3Ω 

R₃ = 3Ω 

• Parallel combination: 

R₁, R₂ and R₃ connected in stair like pattern one above the other.

\(\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\)

\(\frac{1}{R_{eq}}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\)

\(\frac{1}{R_{eq}}=\frac{3}{3}\)

\({R_{eq}}=\frac{3}{3}Ω\)

R_eq = 1Ω

• Series Combination: 

R₁, R₂ and R₃ connected in same line across the potential difference V

R_eq = R₁ + R₂ + R₃

R_eq = 3 + 3 + 3

R_eq = 9 Ω

• Mixed combination: 

R₁ is connected in series with the parallel combination fo R₂ and R₃.

\(\frac{1}{R_p}=\frac{1}{R_2}+\frac{1}{R_3}\)

Putting the values of the resistance, we get 

Thus, we get R_p = 1.5 Ω

R_s = R₁+ R_p = 3 +1.5= 4.5 Ω 

Thus, the net resistance of the circuit is 4.5Ω