When studying for an electrical fundamentals exam or debugging a bench prototype, abstract formulas rarely stick until you apply them to a concrete circuit. A classic example of resistance in electricity involves mixed series-parallel networks. These circuits test your ability to reduce complex topologies into a single equivalent resistance ($R_{eq}$), calculate node voltages, and—crucially for real-world engineering—verify that your components will not overheat and fail. Below, we break down a standard exam problem, showing every algebraic step, identifying common traps, and verifying the results using Kirchhoff’s laws.
Problem Statement
A 24V DC ideal voltage source ($V_s$) is connected to a resistor network. The network consists of a $50\Omega$ resistor ($R_1$) in series with a parallel junction (Node A). Node A splits into two branches:
- Branch 1: A single $60\Omega$ resistor ($R_2$).
- Branch 2: A $20\Omega$ resistor ($R_3$) in series with a $40\Omega$ resistor ($R_4$).
Find: The total equivalent resistance ($R_{eq}$), the total current drawn from the source ($I_{total}$), and the specific voltage drop across $R_4$ ($V_{R4}$).
Methodology: Which Theorem Applies and Why?
For this topology, Series-Parallel Reduction combined with Ohm’s Law and the Voltage Divider Rule is the most efficient method. We use this approach because the circuit contains only independent sources and simple resistive branches. There are no complex bridges (like a Wheatstone bridge) or dependent sources that would necessitate Mesh Current Analysis or Nodal Voltage Analysis. By systematically collapsing series and parallel combinations from the load side back toward the source, we can find $R_{eq}$ and work forward to find specific node voltages.
Step-by-Step Solution: Algebraic Reduction
The most common mistake in this type of problem is attempting to combine parallel branches before resolving the series components hidden inside them. We must work from the furthest point from the source inward.
- Resolve Branch 2 (Series Reduction):
Resistors $R_3$ and $R_4$ are in series with each other. We combine them into a single equivalent resistor for Branch 2 ($R_{B2}$).
$R_{B2} = R_3 + R_4$
$R_{B2} = 20\Omega + 40\Omega = 60\Omega$ - Resolve Node A (Parallel Reduction):
Now, Branch 1 ($R_2 = 60\Omega$) is in parallel with the newly simplified Branch 2 ($R_{B2} = 60\Omega$). We use the product-over-sum formula for two parallel resistors to find the equivalent resistance of the parallel block ($R_p$).
$R_p = \frac{R_2 \times R_{B2}}{R_2 + R_{B2}}$
$R_p = \frac{60 \times 60}{60 + 60} = \frac{3600}{120} = 30\Omega$ - Calculate Total Equivalent Resistance ($R_{eq}$):
The parallel block ($R_p$) is in series with the main line resistor ($R_1$).
$R_{eq} = R_1 + R_p$
$R_{eq} = 50\Omega + 30\Omega = 80\Omega$ - Calculate Total Current ($I_{total}$):
Using Ohm’s Law across the entire circuit.
$I_{total} = \frac{V_s}{R_{eq}}$
$I_{total} = \frac{24V}{80\Omega} = 0.3A \text{ (or } 300mA\text{)}$ - Calculate Voltage at Node A ($V_A$):
The voltage at Node A is the source voltage minus the drop across $R_1$.
$V_{R1} = I_{total} \times R_1 = 0.3A \times 50\Omega = 15V$
$V_A = V_s - V_{R1} = 24V - 15V = 9V$
(Note: You can also find this using the voltage divider rule: $V_A = 24V \times \frac{30\Omega}{50\Omega + 30\Omega} = 9V$). - Calculate Voltage Drop Across $R_4$ ($V_{R4}$):
Node A applies 9V across Branch 2. We use the voltage divider rule within Branch 2 to find the drop specifically across $R_4$.
$V_{R4} = V_A \times \frac{R_4}{R_{B2}}$
$V_{R4} = 9V \times \frac{40\Omega}{60\Omega} = 9V \times \frac{2}{3} = 6V$
Sanity Check and Independent Verification
A correct algebraic derivation means nothing if the physical reality of the circuit is ignored. Let’s run a sanity check on units, order of magnitude, and power dissipation.
Order of Magnitude & Units: $R_{eq}$ must be greater than the series resistor ($50\Omega$) but less than the total sum of all resistors ($170\Omega$). Our answer of $80\Omega$ fits perfectly. All units resolve correctly to Ohms, Amps, and Volts.
Independent Verification (KCL & KVL):
Let’s verify using Kirchhoff’s Current Law (KCL) at Node A. The total current entering Node A is $300mA$.
Current through Branch 1: $I_{R2} = \frac{9V}{60\Omega} = 150mA$.
Current through Branch 2: $I_{B2} = \frac{9V}{60\Omega} = 150mA$.
$150mA + 150mA = 300mA$. KCL holds true.
Kirchhoff’s Voltage Law (KVL) around the outer loop: $24V - V_{R1} - V_{R2} = 24V - 15V - 9V = 0V$. KVL holds true.
The Real-World Power Trap:
In an exam, you stop at the voltage and current. On the workbench, you must check power dissipation. According to HyperPhysics, power is calculated as $P = I^2R$. If you built this using standard 1/4W (0.25W) carbon film resistors, the circuit would fail catastrophically.
| Component | Current | Power Dissipated ($I^2R$) | Standard 1/4W Rating | Required Real-World Rating |
|---|---|---|---|---|
| $R_1$ ($50\Omega$) | 300 mA | 4.50 W | FAIL (Over by 18x) | 5W or 10W Wirewound |
| $R_2$ ($60\Omega$) | 150 mA | 1.35 W | FAIL (Over by 5.4x) | 2W Metal Oxide |
| $R_3$ ($20\Omega$) | 150 mA | 0.45 W | FAIL (Over by 1.8x) | 1W Carbon Film |
| $R_4$ ($40\Omega$) | 150 mA | 0.90 W | FAIL (Over by 3.6x) | 1W or 2W Metal Film |
As shown in the table, $R_1$ is dissipating 4.5 Watts. A standard 1/4W resistor will literally catch fire. When measuring physical resistance on a prototype, always use a high-quality multimeter like the Fluke 87V to verify component values before applying power, and ensure your wattage ratings exceed calculated dissipation by at least 50% for thermal safety.
Frequently Asked Questions
What is a real-world example of resistance in electricity?
A common real-world example of resistance in electricity is the heating element in a toaster or space heater. These devices use high-resistance wire (like Nichrome) specifically designed to convert electrical energy into heat via $I^2R$ losses. Another practical example is a current-limiting resistor placed in series with an LED to prevent it from drawing excessive current and burning out when connected to a voltage source higher than its forward voltage rating.
How do you calculate an example of resistance in electricity with multiple branches?
For multiple branches, you must identify whether the branches are in series or parallel. If they share the exact same two nodes, they are in parallel, and you use the reciprocal formula: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n}$. If the current must flow through one resistor to reach the next, they are in series, and you simply add them: $R_{eq} = R_1 + R_2 + ... + R_n$. For complex, non-series-parallel networks (like a Pi or T network), you must apply Delta-Wye ($\Delta$-Y) transformations or Thevenin’s Theorem to simplify the circuit.
Can you give an example of resistance in electricity causing a voltage drop?
Yes. Wire resistance is a primary cause of voltage drop in residential and industrial wiring. For example, if you run 100 feet of 14 AWG copper wire to a 15A load, the wire itself has resistance (approximately 0.31 ohms per 100 ft for the round-trip loop). Using Ohm’s Law ($V = IR$), the voltage drop is $15A \times 0.31\Omega = 4.65V$. On a 120V circuit, this leaves only 115.35V at the load, which can cause motors to overheat or lights to dim. This is why the NEC recommends keeping voltage drop under 3% for branch circuits.
Why does an example of resistance in electricity change when components heat up?
Resistance is highly dependent on temperature. In standard conductors like copper or carbon film resistors, resistance increases as temperature rises due to increased atomic lattice vibrations scattering electrons. This is defined by the Temperature Coefficient of Resistance (TCR). Conversely, in semiconductors and NTC (Negative Temperature Coefficient) thermistors, resistance drops as they heat up because more charge carriers are freed into the conduction band. This thermal drift is why precision circuits require metal foil resistors with ultra-low TCR ratings (e.g., $\pm 2$ ppm/°C) to maintain accuracy.
