# Review Ohm's Law for Trouble-Shooting Constant Current Source

by Lewis Loflin

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This is a short review of Ohm's Law as it relates constant current sources and series-parallel circuits.

Note: click on any image for larger view.

Fig. 1 is the test setup for a series of articles related to circuit analysis. I assume one has a basic knowledge of Ohm's Law and DC power supplies.

Fig.2 illustrates a basic series-parallel circuit with resistors. For simplicity I used three 10-Ohm resistors.

The amp meter is in series with R3 both are in series with the parallel combination of R1, R2. that together form a 5-Ohm resistor.

Let's note some properties of series-parallel circuits:

In a series circuit current is the same through all elements. The current through R3 equals the current through R1 || R2.

Total resistance is the sum of the individual resistances. Thus R3 = 10-ohms and R1/R2 = 5-ohms for a total of 15-ohms.

The current is 15-volts divided by 15-ohms equal 1 amp.

In a series circuit voltage divides based on the resistance values of each element. The sum of the voltage drops equals the input voltage.

Voltage drop across R3 is 10-ohms X 1 amp = 10-volts. The drop across the 5-ohm combination of R1/R2 is 5-ohms X 1 amp = 5-volts. 5-volts + 10-volts = 15-volts.

In a series circuit power divides based on individual voltage drops. The sum of power in each element adds back to total input power.

With R3 10-volts X 1 amp = 10-watts. With R1/R2 at 5-ohms 5-volts X 1 amp = 5-watts. Total power with all three resistors is 15-watts. 15-volts input X 1 amp = 15-watts.

Note the ammeter depicted in the upper right hand corner. This is a very low resistance device. It presents very little voltage drop and is ignored in these calculations.

Note: ammeters are ALWAYS used in series and never in parallel!

In a parallel circuit such as R1/R2 the voltage across each element is equal. The current will divide based on individual resistance, but add back to total current.

In this case R1 and R2 both equal 10-ohms and the current will divide equally. Going by resistance formula 1 / ( ( 1 / 10 ) + ( 1 / 10) ) = 1 / 0.2 = 5-ohms.

For R1 5-volts / 10-ohms = 500mA. For R2 5-volts / 10-ohms = 500mA. Both add back to 1 amp. Power as current X volts for both produce 2.5 watts of heat. Adds back to 5-watts.

Whew, got that out of the way.

In Fig. 3 one of the parallel 10-ohm resistors has failed producing a circuit with two 10-ohm resistors in series.

The current through both resistors is the same, the voltage drop across each and power is equal, but the total resistance has changed to 20-ohms.

15-volts divided by 20-ohms = 750mA. The voltage drops are now 7.5-volts across each resistor. Power in each resistor is 7.5-volts X 750mA = 5.625 watts. Multiply by two equals 11.25 watts.

Multiply 15-volts input X 750mA = 11.25 watts.

**Fig. 4 Series-parallel circuit connected to constant current source.**

What is a constant current source? One web definition:

A constant current source is a power source which provides a constant current to a load, even despite changes and variance in load resistance.

In Fig. 4 I have connected a constant current source to our three series-parallel resistors. The power input is 18-volts.

Unlike Fig. three when the total resistance changed and change the current, open either R1 or R2 will not change the current from 1 amp I set it for.

The 5-ohm parallel combination will become 10-ohms, but the fixed 1 amp will produce 10-volts on the remaining resistor. The voltage drop will change but not the current.

See Constant Current Source Theory Testing

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