Circuitry problem solving

Everything You Need in One Place

Homework problems? Exam preparation? Trying to grasp a concept or just brushing up the basics? Our extensive help & practice library have got you covered.

Learn and Practice With Ease

Our proven video lessons ease you through problems quickly, and you get tonnes of friendly practice on questions that trip students up on tests and finals.

Instant and Unlimited Help

Our personalized learning platform enables you to instantly find the exact walkthrough to your specific type of question. Activate unlimited help now!

  1. Introduction to Circuitry Problem Solving:
  2. A crash course review on electric circuits.
  3. Concept 1: How are individual resistances related to voltage drops?
  4. Concept 2: What can we conclude about resistors in series?
  5. Concept 3: What can we conclude about resistors in parallel?
  6. Concept 4: How can we determine the brightness of a lightbulb?
  1. Problem Solving for Resistors and Voltage Drops
    The two circuits below are configured as shown. Circuit B includes an additional resistor R3 placed in parallel with R1. Compare the values of VR1 and VR2 in circuit BB compared to circuit AA (no change, decrease, increase)?

    Circuitry Problem Solving
    1. Problem Solving for Lightbulb Brightness with Switches
      Two identical circuits are shown below, except Circuit A has an open switch and Circuit B has a closed switch. How does each lightbulb (R1,R2,R3,R4R_{1}, R_{2}, R_{3}, R_{4} and R5R_{5}) change in brightness compared to when the switch was open? (same, dimmer, or brighter)

      Circuitry Problem Solving
      1. Problem Solving for Complex Circuit with Missing Values

        Circuitry Problem Solving
        1. Find the value of the resistor R4.
        2. Find the voltage (potential difference) of the battery.
        3. Explain how to add one more resistor to make R1 brighter and dimmer.
      2. Solving for Lightbulbs and Terminal Voltage
        The circuit is connected to three identical lightbulbs:

        Circuitry Problem Solving
        1. Find the value of the internal resistance.
        2. Explain whether each lightbulb becomes brighter, dimmer, or the same after opening the switch.
        3. Will the terminal voltage after opening the switch be: greater than, the same as, or less than before opening the switch?
      Topic 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


      • 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=βˆ‘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=βˆ‘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=Vtotalβ‹…RxRtotalV_{x} = V_{total} \cdot \frac{R_{x}}{R_{total}}
        • The concept of terminal voltage and calculations: Vterm=Ο΅βˆ’IrV_{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 currentβ€”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 currentβ€”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)