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84 Direct-current circuit analysis

5-2 Analysis of voltage in a series circuit. See text for discussion.

How do you find the voltage across any particular resistor Rn in a circuit like the one
in Fig. 5-2? Remember Ohm’s Law for finding voltage: E IR. The voltage is equal to the
product of the current and the resistance. Remember, too, that you must use volts,
ohms, and amperes when making calculations. In order to find the current in the circuit,
I, you need to know the total resistance and the supply voltage. Then I E/R. First find
the current in the whole circuit; then find the voltage across any particular resistor.
Problem 5-1
In Fig. 5-2, there are 10 resistors. Five of them have values of 10 Ω, and the other five
have values of 20 Ω. The power source is 15 Vdc. What is the voltage across one of the
10 Ω resistors? Across one of the 20 Ω resistors?
First, find the total resistance: R (10 5) (20 5) 50 100 150 Ω. Then
find the current: I E/R 15/150 0.10 A 100 mA. This is the current through each
of the resistors in the circuit.

If Rn 10 Ω, then En I(Rn) 0.1 10 1.0 V.
If Rn 20 Ω, then En I(Rn) 0.1 20 2.0 V.

You can check to see whether all of these voltages add up to the supply voltage.
There are five resistors with 1.0 V across each, for a total of 5.0 V; there are also five re-
sistors with 2.0 V across each, for a total of 10 V. So the sum of the voltages across the
resistors is 5.0 V 10 V 15 V.
Problem 5-2
In the circuit of Fig. 5-2, what will happen to the voltages across the resistors if one of
the 20-Ω resistors is shorted out?
In this case the total resistance becomes R (10 5) (20 4) 50 80 130
Ω. The current is therefore I E/R 15/130 0.12 A. This is the current at any point
in the circuit. This is rounded off to two significant figures.
The voltage En across Rn 10 Ω is equal to En I(Rn) 0.12 10 1.2 V.