Question

A resistor of $6k\Omega $ with tolerance $10\% $ and another resistance of $4k\Omega $ with tolerance $10\% $ are connected in series. The tolerance of combination is about
A)$5\% $
B)$10\% $
C)$12\% $
D)$15\% $

Answer 1

Hint: Tolerance of a resistor is the maximum error it can have from the original value. The error can be on the positive side or on the negative side. First, we will calculate the error in both of the resistance. When we add the two resistors in series, we can use the fact that error can be directly added in case of addition or subtraction.

Complete step by step answer:
Let the two resistors be $R = 6k\Omega \pm 10\% $and $r = 4k\Omega \pm 10\% $
First, we calculate maximum values of error in both the resistors
Maximum error in $R$, ${e_1} = \dfrac{{10}}{{100}} \times 6 = 0.6k\Omega $
Maximum error in $r$,${e_2} = \dfrac{{10}}{{100}} \times 4 = 0.4k\Omega $
The resistors can be rewritten as,
$R = (6 \pm 0.6)k\Omega $and $r = (4 \pm 0.4)k\Omega $
When these two resistors are connected in series equivalent resistance will become $R + r$
${R_{eq}} = (6 \pm 0.6) + (4 \pm 0.4)$
As we know, error can be directly added in addition and subtraction,
${R_{eq}} = (10 \pm 1)k\Omega $
Error percentage $ = \dfrac{1}{{10}} \times 100 = 10\% $
We can rewrite equivalent resistance as,
${R_{eq}} = 10k\Omega \pm 10\% $
B) is correct.

Additional information:
Error is the difference between the actual value and calculated value of any physical quantity. Basically there are three types of errors in physics, random errors, blunders, and systematic errors.

Note: When you add or subtract two numbers with errors, you just add the errors (you add the errors regardless of whether the numbers are being added or subtracted).