Circuits

V = IR

• Voltage is electric potential. It is measured in volts (V).

• I is current. This is the flow rate of electrical charge. It is measured in Amps (A).

• R is resistance to electric current. It is measured in Ohms (Ω).

Simple ExampleSolve for the current through the resistor.

V = IR12V = I (4 Ω)I = 3A

Resistors in Series

Current is the same through all resistors. If it goes through one resistor, it has to go through all of them.

Total resistance is the sum of all resistances.

Part of the voltage drops across each resistor.

Voltage drop across a resistor is proportional to the resistance.

In the circuit to the right, solve for the current through and voltage drop across each resistor.

R=R1 + R2 = 4Ω + 8Ω = 12Ω

V=IR12V=I(12Ω)I = 1A

Current through this circuit is 1A.

In the first resistor, V=IRV=(1A)(4Ω)V=4V

In the second resistor, V=IRV=(1A)(8Ω)V=8V

Resistors in Parallel

The voltage drop is the same across each resistor in parallel.

1/R = 1/(R1) + 1/(R2)

The current can go through any of them with the same voltage drop.

Current is different in each resistor following Ohm’s Law.

Find the voltage drop across and current through each resistor. Solve for the equivalent resistance.

For both resistors, V=12V.

For the 4Ω resistor, V=IR12V=I(4Ω)I=3A

For the 6Ω resistor, V=IR12V=I(6Ω)I=2A

Equivalent resistance:1/R = 1/R1 + 1/R21/R = ¼ + 1/61/R = 10/24R = 2.4Ω

Check:

Voltage = total current * equivalent resistance12V = (2A + 3A) * (2.4Ω)12V= (5*2.4)V12V=12V

Circuit Reduction

1) Start by adding any resistors in series with each other.

2) Find the equivalent resistance of any resistors in parallel with each other.

3) Repeat as necessary until you have found the equivalent resistance for the entire circuit.

Tip: Start as far from the voltage source as possible and work toward the voltage source.

Tip 2: Redraw the circuit at the beginning to make it easier to work with.

30

1010

24

409

6

12

30

1010

24

4096

12

40

10

24

40918

20

10

24

6

30

30

15

Power

Power dissipate by a resistor is given by the equation P=IV.

You can substitute IR in place of V and get P=I2R.

Another equivalent expression is P=V2/R