COMPLEX CIRCUITS AND POWER

Unit 8: Electricity and Magnetism

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REVIEW: SERIES CIRCUIT

• As the current goes through the circuit, the charges must USE ENERGY to get through the resistor
• So each individual resistor will get its own individual potential voltage).
• We call this VOLTAGE DROP.
• Voltage drop can be calculated by I = V/R using the current (same at all points) and the specific resistor

Note: “equivalent resistance” means TOTAL resistance

V_T = V₁ + V₂ + V₃

R_T = R₁ + R₂ + R₃

I_T = I₁ = I₂ = I₃

REVIEW: PARALLEL CIRCUIT

• In a parallel circuit, we have multiple loops
• The current splits up among the loops with the individual loop currents adding to the total current

• It is important to understand that parallel circuits will all have some position where the current splits and comes back together
• We call these JUNCTIONS.
• The current going IN to a junction will always equal the current going OUT of a junction

V_T = V₁ = V₂ = V₃

I_T = I₁ + I₂ + I₃

1/R_T = 1/R₁ + 1/R₂ + 1/R₃

SERIES VS PARALLEL CIRCUITS

Disadvantages of Series

• When one component fails the whole circuit fails
• Current is the same at all points and the voltage is divided between the bulbs
• The more bulbs added the dimmer each one is

Advantage of Series

• Do not get heated easily

Advantages of Parallel

• When one bulb fails the rest of the circuit continues to work
• Voltage in each branch is the same as the supply voltage, so each bulb in parallel will be as bright as a single bulb

Disadvantages of Parallel

• More wire is needed

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COMPLEX CIRCUITS

Many times you will have series and parallel in the SAME circuit

Part is working in series, and part in parallel

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WHAT IS THE TOTAL RESISTANCE?

parallel

EXAMPLE 1

Calculate the total (equivalent) resistance of the complex circuit.

R1 and R2 work in parallel

1/R_parallel = 1/R1 + 1/R2

1/R_parallel = 1/20 + 1/35

R_parallel = 12.7 Ω

R3 and R1&2 work in series

R_T = R_parallel + R_series

R_T = 12.7 + 4

R_T = 16.7 Ω

EXAMPLE 2

Calculate the total (equivalent) resistance of the complex circuit.

R2 and R3 work in parallel

1/R_parallel = 1/R2 + 1/R3

1/R_parallel = 1/1500 + 1/1500

R_parallel = 750 Ω

R1, R4, and R2&3 work in series

R_T = R_parallel + R_series

R_T = 750 + 100 + 150

R_T = 1000 Ω

ELECTRICAL POWER

ELECTRICAL POWER

• Electrical Power (P) is the rate at which electrical energy is transferred
• The unit for Power (P) is the Watt (W)
• 1 Watt = 1 Joule/Coulomb

P = I V = I²R = V²/R

ELECTRICAL POWER

• P = IV can be used when looking at the overall power of a circuit
• When calculating power for individual resistors, V²/R should be used

P = I V = I²R = V²/R

EXAMPLES

1. A 6 V battery produces a current of 0.5 A. What is the power in the circuit?

P = IV

P = (6)(0.5)

P = 3 W

2. A 100 W light bulb is operating on 1.2 A. What is the voltage?

P = IV

100 = 1.2V

V = 83 V

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EXAMPLES

3. Two 5 Ω lightbulbs are connected in a parallel circuit with a 6 V battery. How much power does each bulb generate?

P = V²/R

P = 6²/5

P = 7.2 W per bulb

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BULB BRIGHTNESS

Lightbulb brightness is determined by power (watts)

BULB BRIGHTNESS

• Three identical lightbulbs are connected in series to a 9V battery.
• If one lightbulb is removed and the other two are reconnected, what happens to the brightness of the remaining bulbs?
• The bulbs will be brighter.
• Power is V²/R. If there are fewer bulbs, there will be more voltage per bulb, resulting in more power.

BULB BRIGHTNESS

• Three identical lightbulbs are connected in parallel to a 9V battery.
• If one lightbulb is removed, what happens to the brightness of the remaining bulbs?
• The bulbs will have the same brightness.
• Power is V²/R. Since voltage is equal across a parallel circuit, each resistor will continue to receive the same voltage and therefore will have the same power.

8.4 LESSON CHECK