Circuitry problem solving - Electric Circuits

Circuitry problem solving

Lessons

Notes:

In this lesson, we will learn:

  • A brief review on voltage, current, and resistance
  • Establishing 4 main concepts for problem solving:
    • Concept #1: a smaller resistor uses up less voltage; a bigger resistor uses up more voltage
    • Concept #2: the more resistors added in series with the battery into the circuit will increase the total equivalent resistance
    • Concept #3: the more resistors added in parallel with the circuit will decrease the total equivalent resistance
    • Concept #4: the brightness of a lightbulb is related to the voltage drop across it (as well as the power dissipated by it)
  • Solving questions for more conceptual electric circuits questions:
    • Using a combination of all previous concepts and formulas (V,I,R,V, I, R, Ohm’s Law, VtermV_{term}, Power)
    • As well as applying the 4 main concepts

Notes:

  • Before facing problem solving questions for electric circuits that are oftentimes just as conceptual as they are mathematical, one must have a firm understanding of the concepts of each lesson thus far:
    • Voltage: staircase analogy, Kirchhoff’s Loop Rule, equal voltage in parallel
    • Current: water analogy, Kirchhoff’s Junction Rule
    • Resistance: calculating total resistance for series vs. parallel configurations
      • Req(series)=R1+R2+R3+...Rn=k=1nRk R_{eq (series) = R_{1} + R_{2} + R_{3} + ... R_{n} = \sum_{k = 1}^n R_{k}}
      • 1Req(parallel)=1R1+1R2+1R3+...1Rn=k=1n1Rk \frac{1}{R_{eq (parallel)}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} + . . . \frac{1}{R_{n}} = \sum_{k = 1}^n \frac{1}{R_{k}}
    • The conceptual relationships as defined by Ohm’s Law: V=IRV=IR
    • The shortcut for Ohm’s law; the voltage divider method: Vx=VtotalRxRtotalV_{x} = V_{total} \cdot \frac{R_{x}}{R_{total}}
    • The concept of terminal voltage and calculations: Vterm=ϵIrV_{term} = \epsilon - Ir
    • Power: total power is additive, P=EtP = \frac{E}{t} and P=IV=I2R=V2RP = IV = I^{2}R = \frac{V^{2}}{R}
    • Energy: E=PtE=P t and E=IVtE=I V t

  • The 4 main concepts can be summarized as follows:
    • I. The greater the resistance of a resistor, the more voltage that it uses up (and vice versa; a smaller resistor uses less voltage)
    • II. The more resistors added in series, the greater the equivalent resistance
    • III. The more resistors added in parallel, the lesser the equivalent resistance
    • IV. The brightness of a lightbulb is related to the voltage it uses up (its voltage drop) as well as, the power dissipated by it

  • The brightness of a lightbulb is related to the amount of voltage that it uses up (voltage drop); the more voltage used, the brighter the light bulb
    • The voltage drop is dependent on current and resistance (V=IRV=IR)
  • The brightness of a lightbulb can also be understood as how hot the filament is burning
    • The incandescent lightbulb is transforming electrical energy into thermal and light energy; the rate of energy transformation is power
    • Power is dependent on voltage (P=IVP=I V) as well as current and resistance (P=I2R=V2RP=I^{2}R= \frac{V^{2}}{R} )
  • When observing lightbulbs in series:
    • Adding more lightbulbs in series will increase the overall resistance, thus diminishing the total current—this leads to a smaller voltage drop across each lightbulb, causing a dimming effect
    • Opening a switch or having a single broken lightbulb in the series chain will cause all relevant lightbulbs turn off (the whole circuit will be compromised)
  • When observing lightbulbs in parallel:
    • Adding more lightbulbs in parallel will decrease the overall resistance, thus increasing the total current—the balance leads to relatively constant brightness across all parallel lightbulbs
    • Opening a switch or having a single broken lightbulb will not compromise the whole circuit; only the relevant branch of the circuit will be affected (turned off)
  • Intro Lesson
    Introduction to Circuitry Problem Solving:
  • 3.
    Problem Solving for Complex Circuit with Missing Values

    Circuitry Problem Solving
  • 4.
    Solving for Lightbulbs and Terminal Voltage
    The circuit is connected to three identical lightbulbs:

    Circuitry Problem Solving
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Circuitry problem solving

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