Discussion Forum

A resistor of 6k$\Omega$  with a tolerance 10% and another resistance of 4$k \Omega$ with a tolerance 10% are connected in series. The tolerance  of the combination is about 

Answer:

Explanation:

In series combination equivalent resistance 

$R=R_{1}+R_{2}$

=6+4=$10k \Omega$

 Error in combinbation

$\triangle R = \triangle R_{1}+\triangle R_{2}$

$=\frac{10}{100} \times 6+ \frac{10}{100} \times 4$

 =$0.6+0.4=1$

$\frac{\triangle R}{R}=\frac{1}{10}=10$%