CURRENT ELECTRICITY:Formulae

S1-3-16 Describe qualitatively the relationship among current, voltage and resistance.

KEY WORDSOhm’s Law

Three key statements of electric circuits:1. Voltage and current are directly related.

Electrons given more energy ( voltage) will move faster ( current)

through the circuit.

2. Voltage and resistance are directly related.3. Current and resistance are inversely related

The opposite is true: decrease voltage (energy/coulomb) will decrease current (coulombs/sec)

• Assumes the resistance of the circuit is unchanged

2. Current and resistance are inversely related.

Electrons will move slower ( current) as move obstacles ( resistance) are placed in

the circuit.

• Assumes the voltage of the circuit is unchanged

Electrons have the same voltage (energy/coulomb) but move slower (coulombs/sec) through the circuit

3. Voltage and resistance are directly related.

Circuits with more obstacles ( resistance) will need more energy ( voltage) to move the coulombs at the same

rate.

• Assumes the current of the circuit is unchanged

The opposite is true: decrease voltage is needed to have same current (coulombs/sec) with a decrease in resistance

As more resistors are added in series, the overall current in the circuit decreases

Notice all bulbs get dimmer with more resistance added to the circuit since slower current also means less

energy per second in the circuit

As more resistors are added in parallel, the overall current in the circuit increases.

Notice all bulbs get brighter with more junctions since more pathway will increase current which also means more

energy per second in the circuit

Ohm’s Law - The voltage in a wire is equal to current multiplied by resistance.

I R

VV = I · R

I = VR

R = V I

Ohm’s Law is the mathematic representation of those three key statements of electric circuits

A bulb that has 2.4 A flowing through it, has a resistance of 16 Ω. Calculate the voltage drop across the load – energy lost by the electrons to light up the bulb.

● ●

+

-

V = I · R

= (2.4 A) · (16 Ω)

= 38.4 V

38.4 V

A 9 V battery is hooked up to a metal wire to make a circuit. If the metal wire has a current of 1.38 A, calculate the resistance in the wire.

● ●

+

-

R = V I

9 V 1.38 A

=

= 6.5 Ω

• Current is the same at all points• Sum of voltage drops over resistors equals

source

Resistance in Series:

• Total resistance is the sum of all resistance:

RT = R1 + R2 + R3 + ...

RT = R1 + R2 + R3

= 8 + 8 + 4 = 20 Ω

I = V = 10 V = 0.5 A R 20 Ω

10 V

What is the current in this circuit if each resistor has the following values: R1 = 8 Ω, R2 = 8 Ω, R3 = 4 Ω

• Voltage will be the same across all branches• Branch current will vary depending on

resistance

Resistance in Parallel:

• Total resistance is the reciprocal sum of the reciprocals of the values:

1 = 1 + 1 + 1 + …RT R1 R2 R3

1 = 1 + 1 + 1 = 1 + 1 + 1 = 1 RT R1 R2 R3 8 8 4 2

RT = 2 Ω

10 V

What is the current in this circuit if each resistor has the following values: R1 = 8 Ω, R2 = 8 Ω, R3 = 4 Ω

I = V = 10 V = 5.0 A R 2 Ω

Notice: current is 10x faster with same 3 resistors in

parallel vs. series

Amount of current depends on the resistance: Easier the path (less Ω), more current passes

Individual currents can be found with V = IR:

I1 = 10 / 8 = 1.25 A

I2 = 10 / 8 = 1.25 A

I3 = 10 / 4 = 2.5 A

10 V

Notice that the branch currents sum to the 5 A total current

CAN YOU ANSWER THESE QUESTIONS?

S1-3-16:How do current, voltage and resistance affect the performance of a circuit?

KEY WORDSOhm’s Law