6.1 Basic circuit analysis and Ohm's law

Ohm's_Law_0### is the foundation of electrical circuit analysis. It links voltage, current, and resistance, allowing us to predict and control electrical behavior. This simple yet powerful relationship is key to understanding how electricity flows through circuits.

Mastering Ohm's law opens doors to more complex circuit analysis. We'll explore series and parallel circuits, power dissipation, and problem-solving strategies. These skills are essential for designing and troubleshooting electrical systems in engineering applications.

Ohm's Law in Circuits

Understanding Ohm's Law

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• Ohm's law establishes direct proportionality between current and voltage in a conductor
• Mathematical expression [V = IR](https://www.fiveableKeyTerm:v_=_ir) represents Ohm's law
  • V represents voltage (volts)
  • I represents current (amperes)
  • R represents resistance (ohms)
• Rearranged forms of Ohm's law solve for different variables
  • $I = V/R$ calculates current
  • $R = V/I$ calculates resistance

Applying Ohm's Law to Simple Circuits

• Simple circuit components include power source (battery), conductor (wire), and load (resistor)
• Ensure consistent units and values when applying Ohm's law to circuit calculations
• Example: Calculate current in a circuit with 12V battery and 4Ω resistor
  • $I = V/R = 12V / 4Ω = 3A$
• Example: Determine resistance in a circuit with 9V battery and 0.5A current
  • $R = V/I = 9V / 0.5A = 18Ω$

Series vs Parallel Circuits

Series Circuit Analysis

• Resistors connect end-to-end in series circuits
• Total resistance in series calculated by summing individual resistances
  • $R_{total} = R1 + R2 + R3 + ...$
• Example: Calculate total resistance of 3Ω, 5Ω, and 2Ω resistors in series
  • $R_{total} = 3Ω + 5Ω + 2Ω = 10Ω$
• Current remains constant throughout a series circuit
• Voltage divides across components proportional to their resistance

Parallel Circuit Analysis

• Resistors connect across the same potential difference in parallel circuits
• Total resistance in parallel calculated using reciprocal sum formula
  • $1/R_{total} = 1/R1 + 1/R2 + 1/R3 + ...$
• Simplified formula for two resistors in parallel
  • $R_{total} = (R1 * R2) / (R1 + R2)$
• Example: Calculate equivalent resistance of 6Ω and 3Ω resistors in parallel
  • $R_{total} = (6Ω * 3Ω) / (6Ω + 3Ω) = 18Ω / 9Ω = 2Ω$
• Equivalent resistance in parallel always less than smallest individual resistance
• Voltage remains constant across parallel components
• Current divides among parallel branches

Complex Circuit Simplification

• Identify series and parallel combinations in complex circuits
• Reduce combinations to equivalent resistances for simplified analysis
• Example: Simplify a circuit with two 4Ω resistors in parallel, connected in series with a 2Ω resistor
  • Parallel combination: $R_{parallel} = (4Ω * 4Ω) / (4Ω + 4Ω) = 2Ω$
  • Total resistance: $R_{total} = R_{parallel} + 2Ω = 2Ω + 2Ω = 4Ω$

Power Dissipation in Circuits

Power Basics in Electrical Circuits

• Power measures rate of energy transfer or conversion in circuits (watts)
• Resistive circuits convert electrical energy to heat energy (power dissipation)
• Basic power formula for DC circuits $P = VI$
  • P represents power (watts)
  • V represents voltage (volts)
  • I represents current (amperes)
• Power directly proportional to square of current and resistance value
• Alternative power formulas derived from Ohm's law
  • $P = I^2R$ uses current and resistance
  • $P = V^2/R$ uses voltage and resistance

Power Dissipation Analysis

• Total power dissipated equals sum of power dissipated by each component
• Example: Calculate power dissipated in a 5Ω resistor with 2A current
  • $P = I^2R = (2A)^2 * 5Ω = 20W$
• Power dissipation crucial for circuit design considerations
  • Component selection based on power ratings
  • Heat management in circuit systems
  • Overall efficiency optimization

Power, Voltage, Current, and Resistance

Relationship Between Electrical Parameters

• Power equation $P = VI$ and Ohm's law derivatives govern relationships
• Choose appropriate formula based on known variables
  • $P = VI$ when voltage and current known
  • $P = I^2R$ when current and resistance known
  • $P = V^2/R$ when voltage and resistance known
• Example: Calculate power in a circuit with 9V source and 3A current
  • $P = VI = 9V * 3A = 27W$

Problem-Solving Strategies

• Identify known variables and required calculations in given problems
• Apply circuit reduction techniques for complex circuit power calculations
• Consider both power supplied by source and dissipated by components
  • Power supplied equals power dissipated in steady-state DC circuits
• Calculate efficiency as ratio of output power to input power
  • Example: LED with 2W output powered by 3W input
    • Efficiency = (Output Power / Input Power) * 100% = (2W / 3W) * 100% = 66.7%
• Ensure component power ratings exceed calculated power dissipation
  • Example: Choose resistor rated above 5W for circuit dissipating 4.5W