Circuitry problem solving

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 ($V, I, R,$ Ohm’s Law, $V_{term}$, 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
• $R_{eq (series) = R_{1} + R_{2} + R_{3} + ... R_{n} = \sum_{k = 1}^n R_{k}}$
• $\frac{1}{R_{eq (parallel)}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} + . . . \frac{1}{R_{n}} = \sum_{k = 1}^n \frac{1}{R_{k}}$
• The conceptual relationships as defined by Ohm’s Law: $V=IR$
• The shortcut for Ohm’s law; the voltage divider method: $V_{x} = V_{total} \cdot \frac{R_{x}}{R_{total}}$
• The concept of terminal voltage and calculations: $V_{term} = \epsilon - Ir$
• Power: total power is additive, $P = \frac{E}{t}$ and $P = IV = I^{2}R = \frac{V^{2}}{R}$
• Energy: $E=P t$ and $E=I V t$


• 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 ($V=IR$)
• 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 ($P=I V$) as well as current and resistance ($P=I^{2}R= \frac{V^{2}}{R}$ )
• 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)