In a series circuit (only one path for current to follow):
The total resistance is equal to the sum of all resistances.
The sum of the voltages across the resistors equals the battery voltage (Kirchhoff's Voltage Law).
This voltage is distributed proportionally among the resistors. For
example, if one resistor is twice the value of another resistor in the
same series circuit, it will have twice the voltage across it. If two
resistors are equal, the voltage across them will be equal.
The current is the same everywhere in the circuit.
The voltage is positive where conventional current enters an impedance.
The voltage is positive where conventional current exits a current source.
Exercises:
1.
What is the voltage across R1? _________
What is the voltage across R2? _________
2.
What is the voltage across R1? _________
What is the voltage across R2? _________
What is the voltage acrossR3? _________
3.
What is the voltage across R1? _________
What is the voltage across R2? _________
4.
The voltage across R2 is 8 volts.
What is the battery voltage? _________
5.
The current flowing through R1 is 800 mA.
What is the current flowing through
R2? ________
6.
How much current is flowing
through R1? _________
Question 1: The
resistors are equal, therefore the voltages across them is equal. These
two voltages must add up to the battery voltage (10 volts). Therefore,
each resistor has 5 volts across it. Question 2: Same as
question 1. The resistors are equal, therefore the voltages are equal,
they must add up to 15 volts. Each resistor has 5 volts across it. Question 3: R2 has twice
the value of R1 so it must have twice the voltage across it. The
voltages must add up to 15 volts. Therefore R1 has 5 volts and R2 has
10 volts. Question 4: R2 has
twice the value of R1, it must have twice the voltage. Therefore R1 has
4 volts across it. The battery voltage must be the sum of the resistor
voltages, which are 8 volts and 4 volts. Therefore the battery voltage
is 12 volts. Question 5: This is a series circuit. The current is the same everywhere. Therefore the current through R2 is 800 mA. Question 6: Don't make
the mistake of dividing 20 volts by 5 ohms. R1 has only a fraction of
the battery voltage across it. R1 has 1/4 of the total resistance so it
has 1/4 of the total voltage. 1/4 of 20 volts is 5 volts. 5 volts
divided by 5 ohms = 1 ampere.
Find the mystery resistor:
For this exercise there are two resistors in series with a battery. The
value of one resistor is known and the other is unknown. The potential
of the battery and the current are known. This exercise is useful for
finding the internal resistance of a battery as well as other problems.
What is the value of the top resistor?
Solution: We have enough information to calculate the total resistance
of the circuit using Ohm's law. We have 4.5 volts and 60 mA of current.
What resistance would result in 60 mA with 4.5 volts? Using Ohm's law
we know the voltage so we divide into it. That's 4.5 volts ÷ 60 mA which
gives us 75 ohms.
Total resistance is 75 ohms
Following the rules for series circuits, the total resistance is the
sum of all individual resistors. Therefore, the mystery resistor is the
total resistance (75 ohms) minus the 50 ohm resistor or 25 ohms.
Find the Mystery Voltage:
What is the voltage at the top of this voltage divider?
The ground symbol represents 0 volts. At the top of the 100k resistor
there are 10 volts. What is the voltage at the top of the 200k resistor?
Solution: The ratio of resistances is 2:1 (200:100). Therefore, there
must be twice as much voltage across the 200k resistor as there is
across the 100k resistor. There are 10 volts across the 100k resistor
so there must be 20 volts across the 200k resistor. The mystery voltage
is the sum of the two voltages, which is 30 volts. (See Kirchhoff’s
Voltage Law)
