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Math in your Head
Or, Doing Useful Work in a Traffic Jam

By Ross Hershberger

Equations for electronics math are available anywhere. If you have the excellent Maker’s Notebook, with its reference section and a calculator, you can plug in numbers and get answers to many of your circuitry questions. So why learn to do such estimates in your head? Because thinking through questions gives you an intuitive feel for them. It helps you quickly sort and refine ideas before committing physical resources to them.

Next time you’re stuck in traffic, instead of getting frustrated, use the down time to work out some simple practice problems. It gets easier and it’s a very useful skill to have. With some practice, you’ll be able to look at a schematic and see important things like power consumption, where the work is getting done in the circuit, and the scale of the physical parts involved.

One of my electronics gurus, Ned Carlson, taught me that electronics is largely a +/- 10% business. In most situations, you don’t need an exact value. Solve the few critical calculations to the required precision and go with two significant digits for everything else.

First up is always Ohm’s Law:

R = E/I
E = I * R
I= E/R

Where :
R = Resistance
E = Voltage
I = Current.

The mnemonic that I learned for this is Eagle, Indian, Rabbit. Eagle (E = voltage) flying above Indian (I = current) and Rabbit (R = resistance) on the ground. Remember this and you will always know that when solving for I, E goes above R, and when solving for R, E goes above I.

Next is the equation for power (P) in Watts:

P = I * E
P/E = I
P/I = E
P = I^2 * R

The mnemonic is PIE.

Handling very large or small numbers in your head:

Multiplying or dividing large or small numbers is easier if you shift the decimal places on both numbers until you have numbers you can deal with. If the two numbers are on the same side of the equation, shift one to the right and the other to the left, like so:

X = .002 * 10,000
X = .02 * 1,000
X = .2 * 100
X = 2 * 10
X = 20

If they’re on opposite sides of the equation, shift them both in the same direction by multiplying or dividing both by 10:

100X = .5
10X = .05
X = .005

Some practical examples:

Example A: Current and power from resistance and voltage.

A 920 Ohm resistor is connected from a 45V circuit node to ground. How much current does the resistor draw and how much power does it dissipate?

First, the current:

I = E/R
I = 45V/920 Ohms

Since 45 is less than 920, the current will be much less than 1 amp.
2 times 45 is 90, too small by a factor of 10
20 times 45 is 900, close to 920.
Therefore, I is about 1/20.
1/20 = 0.050 so the current drawn is roughly 50ma.

For a quick mental check, multiply the current by the resistance to see if you get 45V.

E = 920 Ohms * .05A
Shift decimal places
E = 92 * .5
Half of 92 is 46
Pretty close to 45V, so we didn’t lose a decimal place somewhere by accident.

What about power dissipation in the resistor?

P = I * E
P = 0.05A * 45V
Shift decimal places
P = 0.5 * 4.5
Half of 4.5 is 2.25

How close did we get?

Current calculation error:
I = E / R
45V/920 Ohms = 0.0489A
0.0489 – 0.050 = 0.0011 error
(0.0011/0.050) * 100 = 2.17%

Our head estimate was about 2% off, which is acceptable.

Power calculation error:

P = I * E
0.0489A * 45V = 2.2 Watts actual power
((2.2 – 2.25)/2.2) * 100 = 2.27% calculation error. Right in the ballpark.

Example B: Calculate current flow from power dissipation and high impedance.

40 Watts are dissipated in a 10,000 Ohm impedance. What is the current?

P = (I ^2) * R
40 Watts = (I^2) * 10,000 Ohms
I^2 = 40/10,000
Shift decimal places
I^2 = 4/1,000
I^2 = 0.004 (four thousandths)
Take the square root of both sides
I = sqrt(0.004)

Uh oh. Can’t do square roots of small numbers in our head, so we’re stuck. Or are we? Well, we can factor this, take the square roots of the factors and multiply them to get our answer.

Sqrt(0.004) = sqrt(4) * sqrt(0.001)

That doesn’t look much easier, but it is if you remember that the square root of 1/1000 = about 0.03 (actually it’s 0.0316).

The square root of 4 is 2.
So sqrt(0.004) = 2 * 0.03 = 0.06A.

The actual answer is 0.063A.
The error from our mental calculation is ((0.060 – 0.063)/0.063 * 100) = 4.76%.

If you remember that the square root of 1/1000 is about 0.03 and the square root of 1000 is about 30 you can estimate a lot of tricky math without a calculator.

This is useful because resistances and impedances in the range of 10^4 Ohms and currents in the range of 10^-3 or 10^-4 are common in hobby electronics.

What is the voltage across that 10,000 Ohm impedance with 0.06A flowing through it?

E = I * R
E = 0.06A * 10,000 Ohms
Shift decimal places
E = 0.6 * 1,000
Shift decimal places again
E = 6.0 * 100
E = 600V

If we multiply our estimated current by our estimated voltage we should be back at 40 Watts (P = I * E).

0.06A * 600V = 36 Watts?

What happened? That’s 10% too low. We started with 40 watts and lost 4 in the calculation. The error in estimating the current was increased by multiplying the estimated current by the estimated voltage. Be careful not to carry rough estimates too far in calculations without going back and computing a more exact number.

Many of the electronics calculations that you do with a spreadsheet or a calculator, you can now do to useful levels of precision by simplifying the equations to get numbers you can perform simple math with. As you practice this, you will get a feel for the answers and it becomes more intuitive.

Before buying parts or soldering anything, check your mental calculations with a calculator for accuracy.


Bio: Ross Hershberger was working with hobby computers long before IBM coined the term PC. He worked as a Mainframe Systems Analyst for 20 years before returning to technical electronics as a restorer of vintage audio gear. He is the author of “Econowave Speakers” In MAKE Volume 20 and “Squelette the Bare Bones Amplifier” in Volume 23. He is currently working on an unnamed vacuum tube stereo preamp project.

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Gareth Branwyn

Gareth Branwyn is a freelancer writer and the former Editorial Director of Maker Media. He is the author or editor of a dozen books on technology, DIY, and geek culture, including the first book about the web (Mosaic Quick Tour) and the Absolute Beginner’s Guide to Building Robots. He is currently working on a best-of collection of his writing, called Borg Like Me.


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Comments

  1. Ross Hershberger says:

    The two examples cited in the post are from my bench work. The 45V/920 Ohm combination was a bypass resistor for cathode current in a Fisher X-101-B tube audio amplifier. I wanted to reduce the current load through the tubes and needed to know how much was passing through the resistor.
    The 40 watt/10,000 Ohm scenario was the output stage of a tube amp that I was breadboarding. I needed to know the AC voltage and current so that I could estimate the working power supply voltage and current draw at full power.

  2. NILESH PATADE says:

    sir i like electronics subjects and all componant

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