#### Summary:

• 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)

Series Circuit Exercises