Problem 15: A 12Ω resistor is connected with a parallel combination of 10 resistors each of 200Ω. What will be the net resistance of the circuit?

Solution:

Step: 1 Overview

Parallel Combination of resistors.

See overview of problem 7.

Series combination of resistors.

See overview of problem 8.

Step: 2 Calculation

Above problem states that 10Ω resistors are connected in parallel, each of 200Ω, so their net resistance will be:

1/R = 1/R₁ + 1/R₂ + 1/R₃ + 1/R₄ + 1/R₅ + 1/R₆ + 1/R₇ + 1/R₈ + 1/R₉ + 1/R₁₀
1/R = 1/200 + 1/200 + 1/200 + 1/200 + 1/200 + 1/200 + 1/200 + 1/200 + 1/200 + 1/200
1/R = (1+1+1+1+1+1+1+1+1+1)/200
1/R = 10/200
1/R = 1/20
R = 20Ω

Now this combination is connected in series with 12 Ohms resistor, so net equivalent resistance will be:

R_t = R + 20Ω
R_t = 12 + 20
R_t = 32Ω (Ans)