# Quantitative Aptitude for the Electrical Engineer

Published by Arun Isaac on

Tags: math

This is based on my dissatisfaction with the messed up way time and work problems are handled in quantitative aptitude examination guides. They take a simple idea, threaten you with time constraints and leave you feeling broken and guilty because you couldn't solve them quickly enough.

And having an education in fundamental electrical engineering, I couldn't help but notice the underlying symmetry between

\[ \text{Amount of work} = \text{Speed of work} \times \text{Time taken} \]

and

Ohm's Law, given by \[ V = IR \]

So, this is my proposal. Model the given time and work problem as a very simple resistive network and the problem will solve itself!

**Analogies obtained from the formula**

Amount of work is analogous to Voltage Speed of work is analogous to Current Time taken is analogous to Resistance

A few simple problems I have worked out using this analogy are shown below. Much more are available in the pdf available for download at the end of this page.

**Worker A takes 8 hours to do a job. Worker B takes 10 hours to do the same job. How long should it take both A and B working together, but independently, to do the same job?**

\[ R_a = 8 \] \[ R_b = 10 \] A and B working together is analogous to Ra || Rb \[ R_a || R_b = \frac{40}{9} \; \text{days} \] The parallel resistance makes sense when you think of workers A and B working in parallel. So, their individual speeds of working (currents) add up.

The Ohm's Law analogy also holds good for the pipes-cisterns class of problems, with leaks being modelled as negative resistances. I have shown the working of a simple example below.

**An electric pump can fill a tank in 3 hours. Because of a leak in the tank, it took 3½ hours to fill the tank. If the tank is full, how much time will the leak take to empty it?**

\[ R_a = 3 \] \[ R_a || R_b = \frac{7}{2} \] Solving, \[ R_b = -21 \] Therefore, the leak will take 21 hours to empty the tank.