Question

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

Answer 1

Hint: In series, the equivalent resistance of two resistors increases as it is the sum of resistances. The tolerance of the equivalent resistance in series is the ratio of total error in the measurement of resistance to the equivalent resistance.

Formula Used:
The equivalent resistance of the series combination of two resistors of resistance $$R_1$$ and $$R_2$$ is,
$$R_{eq}=R_1+R_2$$
The tolerance of the resistance is given as,
$$t\mathrm{\%}={\displaystyle \frac{\mathrm{\Delta }R}{R}}\times 100$$
Here, R is the resistance and $$\mathrm{\Delta }R$$ is the error in the resistance.

Complete step by step answer:
We express the resistance of the resistor in terms of its error as follows,
$$R_A=\left(6\pm \mathrm{\Delta }{R}_A\right)\mathrm{\Omega }$$.
And, $$R_B=\left(4\pm \mathrm{\Delta }{R}_B\right)\mathrm{\Omega }$$.

We can express the equivalent resistance of series combination of $$R_A$$ and $$R_B$$ as follows,
$$R_{AB}=\left(R_A+R_B\right)\mathrm{\Omega }\pm \left(\mathrm{\Delta }{R}_A+\mathrm{\Delta }{R}_B\right)\mathrm{\Omega }$$
$$\Rightarrow R_{AB}=10\mathrm{\Omega }\pm \left(\mathrm{\Delta }{R}_A+\mathrm{\Delta }{R}_B\right)\mathrm{\Omega }$$

Now we have to calculate the percentage of tolerance of the equivalence resistance as below,
$$\mathrm{\Delta }R=\left({\displaystyle \frac{\mathrm{\Delta }{R}_A+\mathrm{\Delta }{R}_B}{R_A+R_B}}\right)\times 100$$ …… (1)

We have, the tolerance of resistance $$R_A$$ is,
$${\displaystyle \frac{\mathrm{\Delta }{R}_A}{R_A}}=10\mathrm{\%}$$
$$\Rightarrow \mathrm{\Delta }{R}_A=0.1\times 6\text{k}\mathrm{\Omega }$$
$$\Rightarrow \mathrm{\Delta }{R}_A=0.6\text{k}\mathrm{\Omega }$$

And, $$\mathrm{\Delta }{R}_B=0.4\text{k}\mathrm{\Omega }$$
We can substitute $$\mathrm{\Delta }{R}_A=0.6\text{k}\mathrm{\Omega }$$, $$\mathrm{\Delta }{R}_B=0.4\text{k}\mathrm{\Omega }$$, $$R_A=6\text{k}\mathrm{\Omega }$$ and $$R_B=4\text{k}\mathrm{\Omega }$$ in equation (1).
$$\mathrm{\Delta }R=\left({\displaystyle \frac{0.6k\mathrm{\Omega }+0.4k\mathrm{\Omega }}{6k\mathrm{\Omega }+4k\mathrm{\Omega }}}\right)\times 100$$
$$\Rightarrow \mathrm{\Delta }R={\displaystyle \frac{1}{10}}\times 100$$
$$\Rightarrow \mathrm{\Delta }R=10\mathrm{\%}$$
Therefore, the percentage tolerance in series is 10%.

Therefore, the correct answer is option (B).

Note:The tolerance is the error in the measurement of the resistance. It is always given in percentage. To calculate the tolerance, divide the error in the measurement of resistance by the given resistance. In series, the resistance of the circuit increases while in parallel combination of resistors, the equivalent resistance decreases.