Question 1.7: Given two resistors in parallel, R1 = 3 Ω and R2 = 6 Ω, find...

Applying Equation (1.39),

R_{parallel-equiv}=\frac{product-of-both-resistor-values}{sum-of-both-parallel-equiv} .       (1.39)

R_{parallel-equiv}=3\times6/(3+6)=2  Ω.         (1.40)

Equation (1.39) can be arithmetically expressed as follows:

\frac{1}{R_{parallel-equiv}}=\frac{1}{R_{1}}+ \frac{1}{R_{2}}      (1.41)

R_{parallel-equiv}=\frac{R_{1}.R_{2}}{R_{1}+R_{2}}      (1.42)

whereR_{parallel-equiv} refers to the parallel equivalent resistor of R_1 and R_2. Note that Equations (1.41) and (1.42) are equivalent.

Generalizing from Equation (1.41), the parallel equivalent resistance of n
(where n is an integer) that represents the number of resistors equals

1/R_{parallel-equiv}=1/R_{1}+1/R_{2}+1/R_{3}+….+1/R_{n}.      (1.43)

Upon covering Kirchhoff’s laws in the next section we will justify the computations to find series and parallel equivalent resistors.