Question 1.6: Let us assume that both resistors of the previous example ha...
Let us assume that both resistors of the previous example have ±1% tolerance. What is the series equivalent resistance of both resistors with an error of approximately ±1%?
With ±1% tolerance, it is easy to see that
±1% of 1 MΩ = ±10,000 Ω (1.37)
and
±1% of 1 KΩ = ±10 Ω (1.38)
Since the value of the 1 kΩ resistor is much smaller than 1% of the value of the series equivalent resistor, found in the earlier part of this example to be 1.001 MΩ, the entire value of the small resistor can be neglected. The approximate answer is 1 MΩ ±10,000 kΩ, which is within a ±1% error. Note that a 1% error of 1 MΩ from Equation (1.37) is 10,000 Ω.
Resistors in parallel are those resistors that are connected such that the voltage across all of them is the same. Figure 1.11, depicts “n” resistors in parallel. Note that part (a) and part (b) of the figure represent the exact same circuit.
Given two resistors R_1 and R_2 in parallel, the total parallel equivalent resistance (R_{parallel-equiv} ) is
(R_{parallel-equiv} )=\frac{product-of-both-resistor-values}{sum-of-both-parallel-equiv} . (1.39)
