Q. A resistor of 4 k$\Omega$ with tolerance 10% is connected in parallel with a resistor of 6 kW with tolerance 100%. The tolerance of the parallel combination is nearly

A) 10%
B) 20%
C) 30%
D) 40%
Explanation

$R_{s} = \frac{R_{1}R_{2}}{R_{1} + R_{2}},$ $\frac{\mathrm{\Delta}R_{s}}{R_{s}} \times 100$ $= \frac{\mathrm{\Delta}R_{1}}{R_{1}} \times 100 + \frac{\mathrm{\Delta}R_{2}}{R_{2}} \times 100 + \frac{\mathrm{\Delta}(R_{1} + R_{2})}{R_{1} + R_{2}} = 100$ Now, $\mathrm{\Delta}R_{1} = \frac{10}{100} \times 4$k$\Omega$=0.4k$\Omega,$ $\Delta R_{2} = \frac{10}{100} \times 6$k$\Omega = 0.6k\Omega$ Again, $\frac{\mathrm{\Delta}R_{s}}{R_{s}} \times 100 = \frac{0.4}{4} \times 100 + \frac{0.6}{6} \times 100$ $+ \frac{0.4 + 0.6}{10} \times 100$ $= 10 + 10 + 10 = 30$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.