Power

Learning Objectives

• Define power in the context of electric circuits.
• Describe the relationship between voltage (V) and current (I) and their role in power.
• Use and manipulate the basic power formulas to compute power consumption.
• Compute power in series, parallel, and series-parallel circuits.
• Explain passive sign convention and interpret positive/negative power.
• (Advanced) Recognize power concepts for AC circuits: instantaneous, average, apparent, reactive power, and power factor.

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Quick Definition

Power is the rate of energy transfer — how quickly electrical energy is converted into another form (heat, light, mechanical work, etc.).

• Instantaneous power (time domain): $p(t)=v(t)\,i(t)$
• For steady DC, power is constant and measured in watts (W).
• Energy is the integral of power: $E=\int P\,dt$. For constant $P$, $E=P\cdot t$ (Joules).

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Core Formulas (DC / Instantaneous)

1. Basic:

   $$P = V \times I$$

2. Combine with Ohm's law ($V=I\,R$) to get:

   $$P = I^2 R \qquad\text{and}\qquad P=\frac{V^2}{R}$$

3. For time-varying signals (instantaneous):

   $$p(t) = v(t)\,i(t)$$

   Average power over time $T$:

   $$P_\text{avg} = \frac{1}{T}\int_{0}^{T} p(t)\,dt$$

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Units and Prefixes

• 1 watt (W) = 1 joule/second.
• Common prefixes: mW (10⁻³ W), W, kW (10³ W), MW (10⁶ W).
• Horsepower: 1 hp ≈ 745.7 W.

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Passive Sign Convention

• If current enters the terminal of an element labeled with the positive polarity of voltage, then $P=V\cdot I$ is positive → the element absorbs power (consumes).
• If $P<0$, the element is supplying power (a source).
• Conservation: In any complete circuit, algebraic sum of powers = 0.

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Power in Series and Parallel Circuits

Strategy

1. Simplify the circuit (series/parallel reduction).
2. Find total current/voltage.
3. Work back to find currents/voltages on each element.
4. Compute power ($P=VI$, $I^2R$, or $V^2/R$).
5. Verify conservation of power.

Series Circuits

• Current is the same: $I=I_\text{series}$.
• $P_i = I^2 R_i$.

Parallel Circuits

• Voltage is the same: $V=V_\text{parallel}$.
• $P_i = \dfrac{V^2}{R_i}$.

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Worked Examples

Example 1 — Series Resistors (DC)

12 V source, $R_1=2\ \Omega$, $R_2=4\ \Omega$.

• $R_\text{tot}=6\ \Omega$
• $I=12/6=2\ \text{A}$
• $P_1=I^2R_1=8\ \text{W}$
• $P_2=I^2R_2=16\ \text{W}$
• Source power: $12\cdot2=24\ \text{W}$ ✅ Balanced.

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Example 2 — Parallel Resistors (DC)

12 V source, $R_1=3\ \Omega$, $R_2=6\ \Omega$.

• $I_1=12/3=4\ \text{A}$, $P_1=48\ \text{W}$
• $I_2=12/6=2\ \text{A}$, $P_2=24\ \text{W}$
• Source: $I=6\ \text{A}$, $P=72\ \text{W}$ ✅ Balanced.

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Quick Reference Table

Quantity	Symbol	Formula (DC)
Power	$P$	$P = V \cdot I$
Using Ohm’s law	—	$P = I^2 R$
Using voltage only	—	$P = \dfrac{V^2}{R}$
Energy	$E$	$E = P \cdot t$ (J)

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Practice Problems

Problem 1 (Easy)

A 9 V battery is connected to $R_1=1\ \Omega$ and $R_2=2\ \Omega$ in series.
Find the current and power in each resistor.

Solution:
$R_\text{tot}=3\ \Omega,\quad I=3\ \text{A}$
$P_1=9\ \text{W},\quad P_2=18\ \text{W}$

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Problem 2 (Medium)

24 V source, $R_a=8\ \Omega$, $R_b=12\ \Omega$ in parallel.

Solution:
$I_a=3\ \text{A},\quad P_a=72\ \text{W}$
$I_b=2\ \text{A},\quad P_b=48\ \text{W}$
Source: $I=5\ \text{A},\quad P=120\ \text{W}$

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Summary Checklist

1. Mark polarities and directions.
2. Simplify the circuit.
3. Solve for currents/voltages.
4. Compute power ($VI$, $I^2R$, $V^2/R$).
5. Apply passive sign convention.
6. Verify power balance.