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Electronics Lab
Voltage Dividers and More Complicated Circuits
The voltage divider consists of two or more resistors in series and is a surprisingly useful labora- tory device for providing reduced voltages from a power supply. The voltage divider circuit having three or more resistors is not much harder than the two resistor circuit discussed in the previous lab.
A Three Resistor Voltage Divider
A circuit having three resistors R 1 , R 2 , and R 3 in series is indicated below:
If three resistors are in series , then same current i flow through each resistor. Suppose for example, the voltage source V=12 volts and the resistors have values R 1 = 10, 000 W , R 2 = 20, 000 W., and R 3 = 30, 000 W.
IMPORTANT NOTE: You should repeat this experiment using three resistors in your parts box.)Remem- ber, if you use an ohmmeter to measure the values of the resistors, do NOT have the battery attached while making the measurement otherwise you will get incorrect measurement for the resistance.
Since the three resistors are in series, the total resistance is
RTotal = R 1 + R 2 + R 3 (1)
and numerically you have
RTotal = 10, 000 W + 20, 000 W + 30, 000 W = 60, 000 W (2)
Check that the total resistance of three resistors in series is indeed correct using an ohmmeter with the battery not attached.
The current i in this series circuit is given by Ohm's law as
i = (3)
V
RTotal^ =^
12 Volts 60, 000 W =^ 0.2^ ¥^10
- 3 A
or 0.4 milliamps. You can have Mathematica does the calculation obtaining
12 60 000.
If you place an ammeter in each spot indicated in the circuit below, you should get the value of the current i just calculated.
Actually you can place the ammeter anywhere in the above circuit and you will get the same measured value of the current.
The voltage across each resistor is different for the case of resistors in series. For example, the voltage across resistor R 1 is determined with a voltmeter as indicated below:
It is also easy to calculate the voltage V 1 since we already know the current i in resistor R 1. In general,
V 1 = i R 1 (4)
and for the particular example at hand
The voltage V 3 is calculated as below:
V 3 = i R 3 (8)
since the same current flows through both resistors since they are in series. For the particular example at hand
V 2 = I0.2 ¥ 10 -^3 AM ¥ H30, 000 WL = 6 Volts^ (9)
since using Mathematica we obtain
I0.2 ¥ 10 -^3 M ¥ H30 000L
Notice the sum of the voltages across the resistors
V = V 1 + V 2 + V 3 = 2 V + 4 V + 6 V = 12 V (10)
is just the battery voltage. The battery voltage is "divided" across the three resistors in a manner propor- tional to the value of the resistors. The voltage divider can be used to produce any voltage necessary for lab experiments provided the required voltage is LESS THAN the battery or power supply voltage (less than 12 volts in the example above).
Resistors in Series and Parallel
A circuit having three resistors R 1 , R 2 , and R 3 is indicated below:
The "solutioin" to this circuit problem follows a reduction procedure whereby the original, more compli- cated circuit above is replaced successively by simpler circuits. Two of the resistors R 1 and R 2 are in parallel. If two resistors are in parallel , then the same voltage is across each resistor. The total resis- tance of two resistors in parallel is given by
RParallel^ =^
R 1 +^
R 2
Suppose the same two resistors as before, R 1 = 30, 000 W and R 2 = 50, 000 W are in parallel, and this combination is in series with R 3 = 20, 000 W Then the total (or effective) resistance RP of the two resistors in parallel is
RP
30, 000 W
50, 000 W
Using Mathematica we obtain
1 30 000.
and thus
RP = 1 (13)
= 18, 868 W
since Mathematica yields
The procedure now is to reverse the reduction process above and step by step reconstitute the original circuit. Along the way all the currents and voltages are calculated as shown below. The next step is to return to the two resistors in series and calculate the voltage across each:
The voltage across R 3 is given by Ohm's law as
V 3 = i R 3 = 0.31 ¥ 10 -^3 A ¥ 20, 000 W = 6.2 V^ (17) since Mathematica gives
0.31 ¥ 10 -^3 ¥ 20 000.
Similarly, the voltage across RP is given by
VP = i RP = 0.31 ¥ 10 -^3 A ¥ 18, 868 W = 5.8 V^ (18) since Mathematica gives
0.31 ¥ 10 -^3 ¥ 18 868.
Notice that the sum of the voltages 6.2 V+5.8 V = 12 V is the battery voltage. Next return to the original circuit
and notice the voltage across the parallel combination of resistors is just 6.2 Volts just calculated. It follows from Ohm's law that the current through resistor R 1 is given by
i 1 = (19)
5.8 Volts R 1
5.8 Volts 30, 000 W
= 0.19 ¥ 10 -^3 A = 0.19 ma
since Mathematica yields
5. 30 000
This current is measured with an ammeter in series with resistor R 1. Similarly it follows from Ohm's law that the current through resistor R 2 is given by
i 2 = 5.8 Volts (20) R 2
= 5.8 Volts 50, 000 W
= 0.12 ¥ 10 -^3 A = 0.12 ma
since Mathematica yields
5. 50 000
This current is measured with an ammeter in series with resistor R 2. Notice that the sum of the currents through R 1 and R 2 is just 0.19 ma+0.12 ma=0.31 ma which is the same as the current supplied by the battery.
