Circuits Notes

Definition of Potential Difference
- A potential difference, denoted as $
  abla V$ or sometimes represented with a curvy symbol $ ext{e}$, is supplied by batteries, referred to as EMF (Electromotive Force).
- Chemical Energy to Potential Energy
- Batteries convert chemical energy into potential energy when charges ascend a 'charge escalator.'
- Energy Flow in Circuit
- Charges lose potential energy as they traverse the circuit, gaining kinetic energy in the process.
- The energy is eventually dissipated as heat through components like resistors or light bulbs.
- Example: A light bulb connected to a battery releases energy as heat.

Definition of Power
- Power is defined as the rate of energy transfer, calculated as:
 $P = I imes abla V$
 where $I$ is the current, and $
 abla V$ is the potential difference.
- An alternate equation for current is derived as $I = rac{
 abla Q}{
 abla t}$, which connects charge transfer to current.
Units of Power
- The unit for power is watts (W), where 1 watt is equivalent to 1 joule per second $(1 ext{ W} = 1 rac{ ext{J}}{ ext{s}})$.
- It can also be expressed as:
  $1 ext{ W} = 1 ext{ A} imes 1 ext{ V}$
  where amperes (A) relate to current and volts (V) relate to potential difference.

Example Problem: Using Power Equation
- When calculating power, for instance, with $
  abla V = 12 V$ and $I = 3.2 A$, the power calculated:
  $P = 12 ext{ V} imes 3.2 ext{ A} = 38.40 ext{ W}$

Energy Transformation Overview
- A single battery and resistor create a clear pathway of energy transformation:
1. Chemical energy in the battery converts to potential energy.
2. Potential energy turns into kinetic energy while traversing the circuit.
3. Kinetic energy is dissipated as thermal energy in resistors or bulbs through atomic collisions.
Ohm's Law
- The relationship between voltage, current, and resistance in an electrical system is expressed via Ohm's law:
 $ abla V = I imes R$
 Rearranging allows for power calculation in resistors via
 $P = I^2R$

Resistance of 40-Watt Headlight
- Problem: Find the resistance of a 40 watt headlight powered by 12 volts.
- Start with:
 $P = I imes abla V$
- Given $P = 40W$ and $
 abla V = 12V$. Rearranging gives:
 $I = rac{P}{ abla V} = rac{40 W}{12 V} = 3.33 A$
- Solving for resistance now:
 $R = rac{P}{I^2} = rac{40 W}{(3.33 A)^2} = 3.6 ext{ ohms}$
- Unit for resistance is ohm, denoted with the Greek letter omega (Ω).

Comparison of Light Bulbs
- When comparing a 5 watt and 10 watt light bulb, discuss how both maintain the same voltage.
- Understanding that higher wattage correlates with higher current when the potential difference is constant:
- $I$ for 10 watt > $I$ for 5 watt.

Energy vs. Power
- When billing for electricity, companies charge for energy, calculated as:
  $E = P imes t$
- Where energy ($E$) is in kilowatt hours (kWh).
Calculating Monthly Energy Cost
- Example: An electric heater connected to a 120V source drawing 15A.
1. Calculating Power:
 $P = I imes abla V = 15 A imes 120 V = 1800 W = 1.8 kW$
2. Total usage over a month (assuming 3 hours/day for 30 days):
 $ext{Total Energy} = 1.8 kW imes (3 hours/day imes 30 days) = 162 kWh$
3. Monthly Cost Calculation:
 - With a rate of 9.2 cents per kWh:
 ext{Cost} = 162 kWh imes 0.092 $/kWh = 14.92 ext{ dollars}

Direct Current (DC)
- In DC, current flows steadily in one direction, typical in circuits powered by batteries.
Alternating Current (AC)
- AC current reverses direction multiple times per second, often visualized in a sinusoidal wave. The voltage supply can be expressed mathematically:
  $V = V_0 imes ext{sin}(2 ext{π}ft)$
- Key Terms in AC:
- $V_0$ = Peak Voltage
- $f$ = Frequency (oscillations per second)
- $ au$ (lowercase t in function) = Time
- $T$ = Period (time taken to complete one full cycle), related by:
  $T = rac{1}{f}$
- Current in AC follows Ohm's law, showing a similar sinusoidal pattern as voltage during its alternation.