Activity 1.2 4 Circuit Calculations Answer Key

Activity 1.2: 4‑Circuit Calculations – Answer Key and Detailed Walkthrough

In many introductory electrical engineering courses, students are tasked with solving a set of small circuits that involve resistors, voltage sources, and current sources. On top of that, 2 is a classic example that tests one’s ability to apply Ohm’s law, Kirchhoff’s Voltage Law (KVL), Kirchhoff’s Current Law (KCL), and the superposition principle. Consider this: activity 1. In practice, below you will find a complete answer key, step‑by‑step explanations for each of the four circuits, and a concise summary of the key concepts that were used. This guide is designed to help you understand why each answer is correct, not just what the answer is.

Introduction

The four circuits in Activity 1.2 are deliberately simple yet rich enough to cover most of the fundamental techniques used in DC circuit analysis.
Each circuit contains:

1. Resistors (values given in ohms)
2. Voltage sources (DC)
3. Current sources (DC)

The tasks usually ask for:

• The current through a particular resistor
• The voltage across a particular resistor
• The total power dissipated in the circuit

Let’s dissect each circuit one by one Small thing, real impact..

Circuit 1 – Series‑Parallel Combination

Problem Statement

A 12 V battery is connected to a series–parallel network consisting of:

• R₁ = 4 Ω in series with
• a parallel branch of R₂ = 6 Ω and R₃ = 12 Ω.

Find:

1. 3. The voltage drop across R₂. And the total current supplied by the battery. 2. The power dissipated in R₃.

Step‑by‑Step Solution

1. Combine the parallel resistors first.
   [ R_{23} = \frac{R_2 R_3}{R_2 + R_3} = \frac{6 \times 12}{6 + 12} = \frac{72}{18} = 4,\Omega ]

2. Add the series resistor.
   [ R_{\text{total}} = R_1 + R_{23} = 4 + 4 = 8,\Omega ]

3. Apply Ohm’s law to find the total current.
   [ I_{\text{total}} = \frac{V}{R_{\text{total}}} = \frac{12}{8} = 1.5,\text{A} ]

4. Voltage drop across the parallel branch (same as across R₁ because they are series).
   [ V_{23} = I_{\text{total}} \times R_{23} = 1.5 \times 4 = 6,\text{V} ]

5. Voltage across R₂ (same as across R₃ because they are parallel).
   [ V_{R2} = V_{23} = 6,\text{V} ]

6. Power in R₃.
   [ P_{R3} = \frac{V_{R3}^2}{R_3} = \frac{6^2}{12} = \frac{36}{12} = 3,\text{W} ]

Answers (Circuit 1)

Circuit 2 – Voltage Divider with a Load

Problem Statement

A 10 V supply is connected to a voltage divider consisting of R₁ = 2 kΩ and R₂ = 3 kΩ. A load resistor R_L = 1 kΩ is connected across R₂. Determine:

1. The voltage at the midpoint of the divider (node between R₁ and R₂).
2. The current through R_L.
3. The total power delivered by the source.

Step‑by‑Step Solution

1. First, treat the load as a single equivalent resistance in parallel with R₂.
   [ R_{2L} = \frac{R_2 R_L}{R_2 + R_L} = \frac{3,k \times 1,k}{3,k + 1,k} = \frac{3}{4},k = 0.75,k\Omega ]

2. Now the divider is simply R₁ in series with R₂L.
   [ R_{\text{total}} = 2,k + 0.75,k = 2.75,k\Omega ]

3. Total current from the source.
   [ I_{\text{source}} = \frac{V_{\text{source}}}{R_{\text{total}}} = \frac{10}{2.75} \approx 3.636,\text{mA} ]

4. Voltage at the midpoint (node between R₁ and R₂L).
   [ V_{\text{mid}} = I_{\text{source}} \times R_{2L} = 3.636,\text{mA} \times 0.75,k = 2.727,\text{V} ] (Rounded to 2.73 V)

5. Current through the load.
   [ I_{R_L} = \frac{V_{\text{mid}}}{R_L} = \frac{2.727}{1,k} = 2.727,\text{mA} ]

6. Total power delivered by the source.
   [ P_{\text{source}} = V_{\text{source}} \times I_{\text{source}} = 10 \times 3.636,\text{mA} = 36.36,\text{mW} ]

Answers (Circuit 2)

Circuit 3 – Wheatstone Bridge

Problem Statement

A Wheatstone bridge is formed by four resistors: R₁ = 100 Ω, R₂ = 150 Ω, R₃ = 200 Ω, R₄ = 300 Ω. The bridge is powered by a 12 V source across terminals A–C. Find:

1. The voltage difference between the bridge nodes B and D (i.e., V_BD).
2. The current through the galvanometer (assume it is ideal and negligible resistance).

Step‑by‑Step Solution

1. Determine the voltage at node B.
   Node B is the junction of R₁ and R₂.
   The voltage drop across R₁ (from A to B) is: [ V_B = V_{\text{source}} \times \frac{R_1}{R_1 + R_2} = 12 \times \frac{100}{100 + 150} = 12 \times \frac{100}{250} = 4.8,\text{V} ]

2. Determine the voltage at node D.
   Node D is the junction of R₃ and R₄.
   The voltage drop across R₃ (from A to D) is: [ V_D = V_{\text{source}} \times \frac{R_3}{R_3 + R_4} = 12 \times \frac{200}{200 + 300} = 12 \times \frac{200}{500} = 4.8,\text{V} ]

3. Voltage difference V_BD.
   [ V_{BD} = V_B - V_D = 4.8 - 4.8 = 0,\text{V} ]

4. Current through the galvanometer.
   Since V_BD is zero, an ideal galvanometer (zero resistance) draws no current.
   [ I_{\text{galv}} = 0,\text{A} ]

Answers (Circuit 3)

Circuit 4 – Thevenin Equivalent

Problem Statement

A network consists of a 9 V source in series with R₁ = 3 kΩ and a parallel branch of R₂ = 6 kΩ and R₃ = 9 kΩ. Calculate the Thevenin voltage (V_th) and Thevenin resistance (R_th) seen from the terminals across the parallel branch (between points X and Y). Then, if a load R_L = 12 kΩ is connected across X–Y, find the load current.

Step‑by‑Step Solution

1. Find V_th (open‑circuit voltage).
   First compute the equivalent resistance of the parallel branch: [ R_{23} = \frac{R_2 R_3}{R_2 + R_3} = \frac{6 \times 9}{6 + 9} = \frac{54}{15} = 3.6,k\Omega ] Now the circuit is a simple series of 3 kΩ and 3.6 kΩ.
   The voltage across the parallel branch (V_th) is the voltage drop across R_{23}: [ V_{\text{th}} = V_{\text{source}} \times \frac{R_{23}}{R_1 + R_{23}} = 9 \times \frac{3.6}{3 + 3.6} = 9 \times \frac{3.6}{6.6} \approx 4.909,\text{V} ] (Rounded to 4.91 V)

2. Find R_th (short the source).
   With the 9 V source shorted, the remaining resistors are R₁ in parallel with the series of R₂ and R₃:

   • Series of R₂ and R₃: [ R_{23_series} = R_2 + R_3 = 6 + 9 = 15,k\Omega ]
   • Parallel with R₁: [ R_{\text{th}} = \frac{R_1 R_{23_series}}{R_1 + R_{23_series}} = \frac{3 \times 15}{3 + 15} = \frac{45}{18} = 2.5,k\Omega ]

3. Load current with R_L = 12 kΩ.
   Total resistance seen by the load: [ R_{\text{total}} = R_{\text{th}} + R_L = 2.5 + 12 = 14.5,k\Omega ] Load current: [ I_L = \frac{V_{\text{th}}}{R_{\text{total}}} = \frac{4.91}{14.5} \approx 0.339,\text{mA} ]

Answers (Circuit 4)

FAQ – Common Pitfalls & Tips

Conclusion

Activity 1.In real terms, 2 is a micro‑cosm of DC circuit analysis. By systematically applying Ohm’s law, KVL, KCL, and the concept of equivalent resistances, you can solve a wide variety of problems Worth keeping that in mind..

• Always simplify the network step by step (parallel → series).
• Keep track of node voltages; they are the foundation for current calculations.
• Check your work by confirming that power supplied equals power dissipated (energy conservation).

Mastering these four circuits will give you a solid footing for tackling more complex networks, such as those involving dependent sources or transient analysis. Happy calculating!

While the core Activity 1.2 exercises and initial takeaways provide a strong foundation, DC circuit analysis extends far beyond these basic Thevenin equivalent calculations. The following sections explore complementary theorems, real-world constraints, and practical decision-making frameworks to round out your toolkit Easy to understand, harder to ignore..

Norton Equivalents: The Dual of Thevenin’s Theorem

Every linear circuit with a Thevenin equivalent also has a matching Norton equivalent, which swaps the series voltage source for a parallel current source. The two are related by simple source transformation rules:

• Norton current: $I_N = \frac{V_{th}}{R_{th}}$
• Norton resistance: $R_N = R_{th}$ (identical to the Thevenin resistance)

Using the Circuit 4 values derived earlier ($V_{th} \approx 4.That's why 5,k\Omega$ in parallel with the current source. Still, 964 \times \frac{2. Day to day, to verify, calculate the load current for $R_L = 12,k\Omega$ using the Norton model: current divides between $R_N$ and $R_L$, so $I_L = I_N \times \frac{R_N}{R_N + R_L} = 1. 5}{14.Worth adding: 339,\text{mA}$, matching the earlier Thevenin result exactly. 964,\text{mA}$, with $R_N = 2.On top of that, 5,k\Omega$), the Norton equivalent current is $I_N = \frac{4. 5} \approx 1.91,\text{V}$, $R_{th} = 2.91}{2.5} \approx 0.This duality is useful for circuits where parallel combinations are easier to simplify than series networks.

Source Transformation: A Streamlined Simplification Tool

Source transformation is the general process of converting between voltage sources in series with resistance and current sources in parallel with resistance, and it works for both independent and dependent sources (as long as the resistance value remains unchanged). To give you an idea, the original 9 V independent source in series with $R_1 = 3,k\Omega$ in Circuit 4 can be transformed into a 3 mA current source ($9,\text{V} / 3,k\Omega$) in parallel with $3,k\Omega$. This parallel combination can then be combined with other parallel branches in the circuit to reduce the network to a single current source and parallel resistance, which is exactly the Norton equivalent derived above.

This technique is particularly powerful for circuits with multiple branches: instead of solving systems of equations, you can iteratively transform sources until the network is simplified to a single Thevenin or Norton equivalent. It also helps validate your work: if source transformation and Thevenin/Norton calculations yield the same result, you can be confident your equivalent circuit is correct.

Real-World Constraints for Equivalent Circuits

All calculations in Activity 1.2 assume ideal components: resistors with exact values, infinite internal resistance for voltage sources, and zero tolerance for all parts. Real-world circuits deviate from these ideals in predictable ways:

1. Component tolerance: Most commercial resistors have a tolerance rating (±1%, ±5%, or ±10% for common carbon film parts). For Circuit 4, a ±5% tolerance on $R_{th} = 2.5,k\Omega$ means the actual equivalent resistance could range from 2.375 kΩ to 2.625 kΩ, shifting the load current by roughly ±3% even if all other values are exact.
2. Temperature effects: Resistors have a temperature coefficient (e.g., 100 ppm/°C for metal film resistors), meaning their value changes with operating temperature. A 50°C temperature rise would shift a 2.5 kΩ resistor by 1.25%, further altering equivalent circuit values.
3. Non-ideal sources: Real voltage sources (including batteries and regulated power supplies) have internal resistance, which is exactly what the Thevenin $R_{th}$ models for a source network. For a 9 V battery with 1.5 kΩ internal resistance, the Thevenin equivalent would be 9 V in series with 1.5 kΩ, and you can use the same load current calculations to predict how battery voltage sags under load.
4. Dependent sources: Circuits with dependent sources (e.g., transistors or operational amplifiers) require a modified approach to find $R_{th}$: you cannot simply short independent sources, because dependent sources are controlled by circuit variables. Instead, inject a test voltage $V_{test}$ across the open load terminals, calculate the resulting current $I_{test}$, and set $R_{th} = V_{test}/I_{test}$. This same test source method works for any linear circuit, even those without independent sources.

When to Use Thevenin/Norton vs Other Analysis Methods

A clear decision-making rubric helps select the most efficient analysis technique for a given problem:

• Thevenin/Norton: Best when you need to calculate load current or voltage for multiple different load values, or when the load is connected to a fixed "source network" that you do not want to re-analyze each time. It is also the preferred method for modeling non-ideal sources like batteries.
• Mesh (loop) analysis: Most efficient for circuits with few independent loops and mostly voltage sources. It reduces the problem to solving a system of equations equal to the number of loops.
• Node voltage analysis: Best for circuits with few independent nodes and mostly current sources. It avoids dealing with loop currents and works well for circuits with many parallel branches.
• Superposition: Useful for circuits with 2–3 independent sources where you want to avoid solving systems of equations, but it becomes tedious for circuits with many sources, as you must analyze the circuit once per source.

Final Conclusion

The transition from basic Ohm’s law calculations to equivalent circuit theorems like Thevenin and Norton marks a shift from solving individual component values to modeling entire network behaviors. While Activity 1.2 focuses on ideal DC networks, the extensions outlined above—dual Norton equivalents, source transformation, real-world tolerance considerations, and method selection—prepare you to handle practical, non-ideal circuits encountered in real engineering work.

Remember that no single analysis method is universally best: the most efficient approach depends on the circuit topology, the number of sources, and the specific values you need to calculate. Worth adding: as you practice, you will develop an intuition for which tool to reach for first. Whether you are designing a battery-powered sensor, troubleshooting a faulty amplifier, or modeling a complex power distribution network, the foundational principles of linear circuit analysis will remain consistent. Keep experimenting with different methods, validate your results against multiple approaches, and always account for real-world component limitations in your final designs Turns out it matters..

With these skills in hand, you are ready to move beyond static DC networks into AC analysis, transient response, and active circuit design—core topics for any electrical engineering or electronics curriculum.