Parallel Resistors: Why Resistance Decreases

Hello! Welcome to your next lesson in our rapid tour of DC circuits.

In our last lesson, we established that when resistors are in series, they act as a single, longer resistor. Every charge has to pass through every obstacle, so their resistances simply add up, increasing the total opposition to current.

Today, we're going to explore the opposite scenario: resistors in parallel. As previewed, the behavior here is counter-intuitive at first glance. Adding more resistors in parallel actually decreases the total resistance. Your goal for this lesson is to explain physically why this happens and derive the formula for the equivalent resistance of a parallel combination. Mastering this is crucial for simplifying complex networks and deepening your physical intuition for how current behaves when it has choices.

1. The Physical Intuition: More Paths, Less Resistance

Let's start with the "why," as this is the foundation for everything else. The core idea of a parallel circuit is that it provides multiple paths for the current to flow between two common points.

Imagine a crowded store with only one checkout counter open. The line is long, and the rate at which people can leave is slow. The counter is a point of high "resistance." Now, what happens if the manager opens three more counters? The total number of people checking out per minute increases dramatically. Even though you've added more "resistors" (counters), the overall opposition to the flow of people has gone down because you've provided more pathways.

This is precisely what happens in a parallel circuit. Each new resistor you add in parallel is like opening another lane on a highway. It provides an additional route for charge to travel, so the total current that can flow for a given voltage "push" increases. An increase in total current for the same voltage implies a decrease in total resistance.

2. A More Formal Intuition: Conductance

Your instinct to connect resistance to the physical properties of a material ($R=\rho L/A$) is spot on. Adding a resistor in parallel is analogous to increasing the total cross-sectional area ($A$) available for the current to flow through. Since resistance is inversely proportional to area, increasing the effective area decreases the total resistance.

To formalize this, it's incredibly helpful to stop thinking about resistance (how hard it is for current to flow) and start thinking about its reciprocal, conductance (how easy it is for current to flow).

• Resistance (R), measured in Ohms ($\Omega$).
• Conductance (G), measured in Siemens ($S$).
• Relationship: $G=1/R$

With this concept, the logic of parallel circuits becomes simple and elegant. If you have several parallel paths, the total "easiness" of flow is just the sum of the individual "easinesses."

As the article shows, if total conductance is the sum of individual conductances:

$$G_T=G_1+G_2+G_3+\dots$$

And since $G=1/R$, we can substitute this back in:

$$\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\dots$$

This gives us the famous formula for parallel resistance. We've arrived at it simply by reframing the problem in terms of "easiness" of flow instead of "opposition."

3. Deriving the Formula from First Principles

While the conductance approach provides great intuition, a formal proof must be grounded in our fundamental circuit laws: Kirchhoff's Laws and Ohm's Law. Let's derive the formula from scratch.

A video is often best for seeing this step-by-step.

Let's summarize the derivation you just saw. Consider a voltage source $V$ connected across several parallel resistors.

1. Apply KCL: The total current $I_T$ that leaves the source must equal the sum of the currents in each branch.

   $$I_T=I_1+I_2+I_3+\dots$$

2. Apply Ohm's Law: The voltage $V$ across each resistor is identical. We can express the current in each branch using Ohm's Law.

   $$I_1=\frac{V}{R_1},I_2=\frac{V}{R_2},I_3=\frac{V}{R_3}$$

   The total current is related to the equivalent resistance $R_{eq}$ by $I_T=V/R_{eq}$.

3. Substitute and Solve: Substitute these expressions back into the KCL equation:

   $$\frac{V}{R_{eq}}=\frac{V}{R_1}+\frac{V}{R_2}+\frac{V}{R_3}+\dots$$

   Since $V$ is the same everywhere, we can divide it out from every term. This is the crucial step that leaves us with the reciprocals:

   $$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\dots$$

This confirms our result from the conductance approach, but this time it's built directly from the fundamental conservation laws of charge (KCL) and energy (constant voltage).

4. Application and Key Insights

Let's solidify this with a practical example and highlight two powerful rules of thumb.

From the example in the video (a 16V source with 20 Ω and 5 Ω resistors in parallel), we can draw two critical conclusions:

1. The equivalent resistance is always less than the smallest individual resistance.
   In the example, $R_{eq}=4\Omega$, which is less than the smallest resistor, $5\Omega$. This makes perfect sense: if you already have a 5 Ω path and you add any other path (even a high-resistance 20 Ω one), you've given the current an extra option, which can only make the total flow easier (i.e., lower the total resistance).

2. Current divides inversely with resistance.
   The 5 Ω resistor (path of least resistance) drew $3.2A$, while the 20 Ω resistor (path of most resistance) drew only $0.8A$. This confirms the "path of least resistance" intuition. The total current of $4A$ splits, but not evenly.

A common special case is for just two resistors in parallel. The formula can be rearranged to the "product over sum" form, which is often faster to calculate:

$$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{R_2+R_1}{R_1{R}_2}⟹R_{eq}=\frac{R_1{R}_2}{R_1+R_2}$$

For the video's example: $R_{eq}=\frac{20\times 5}{20+5}=\frac{100}{25}=4\Omega$.

Conclusion

Today we've unraveled the seemingly strange behavior of parallel resistors. By thinking in terms of "easiness of flow" (conductance) and applying our fundamental circuit laws, the concepts become clear and logical.

Key Takeaways:

• Physical Cause: Adding resistors in parallel provides more paths for current, which is analogous to increasing a conductor's cross-sectional area. This reduces the overall opposition to flow.
• Conductance View: In parallel, the total conductance is the simple sum of the individual conductances ($G_T=\sum G_i$). This is the most intuitive way to think about it.
• Mathematical Derivation: The formula $1/R_{eq}=\sum 1/R_i$ is derived by applying KCL (currents sum) and Ohm's Law to a parallel configuration where voltage is constant across all branches.
• Practical Rules: The equivalent resistance is always smaller than the smallest parallel resistor, and current preferentially flows through the paths of lower resistance.

Preview of the Next Lesson:
We have now established the rules for series and parallel combinations. In the next two lessons, we will build on this to create powerful analytical shortcuts. We will first derive the Voltage Divider Rule for series circuits, which allows us to find the voltage across any resistor without first calculating the current. Then, we will derive the Current Divider Rule for parallel circuits.