Activity 1.2 – Four‑Circuit Calculations: Answers and Insights
When students tackle Activity 1.2, they are usually asked to analyze four distinct electrical circuits, each with its own set of resistors, voltage sources, and sometimes current sources. The goal is to practice the fundamentals of series‑parallel combinations, Kirchhoff’s voltage and current laws, and Ohm’s law. Below, we walk through each circuit step by step, present the final numerical answers, and explain the reasoning behind every calculation. By the end of this article, you should be able to solve similar problems confidently and understand the underlying concepts that make the solutions work Less friction, more output..
Introduction
In Activity 1.2, four circuits are presented, each a variant of the classic “four‑resistor” problem. Students are required to:
- Determine the equivalent resistance of the entire network.
- Calculate the voltage drop across a specific resistor or branch.
- Find the current flowing through a particular component.
These tasks reinforce the idea that any complex network can be reduced to a single equivalent resistance, after which Ohm’s law can be applied in a straightforward manner.
Circuit 1 – Series‑Parallel Combination
Description
- Resistors: R₁ = 5 Ω, R₂ = 10 Ω, R₃ = 15 Ω, R₄ = 20 Ω
- Voltage source: V = 12 V
- Configuration: R₁ and R₂ are in series; this pair is in parallel with R₃; the resulting network is in series with R₄.
Equivalent Resistance
- Series part: R₁ + R₂ = 5 Ω + 10 Ω = 15 Ω
- Parallel part:
[ \frac{1}{R_{12,3}} = \frac{1}{15 Ω} + \frac{1}{15 Ω} = \frac{2}{15 Ω} ] So, ( R_{12,3} = 7.5 Ω ). - Add R₄: ( R_{\text{eq}} = 7.5 Ω + 20 Ω = 27.5 Ω ).
Total Current
Using Ohm’s law, ( I = V / R_{\text{eq}} = 12 V / 27.Think about it: 5 Ω ≈ 0. 436 A ).
Voltage Across R₃
Since R₃ is in parallel with the 15 Ω series pair, the voltage across R₃ equals the voltage across that pair:
[ V_{R_{12}} = I \times 15 Ω = 0.436 A \times 15 Ω ≈ 6.54 V ]
Thus, the voltage across R₃ is 6.54 V.
Circuit 2 – Two Parallel Branches
Description
- Resistors: R₁ = 8 Ω, R₂ = 12 Ω, R₃ = 24 Ω
- Voltage source: V = 24 V
- Configuration: Branch A contains R₁ in series with R₂; Branch B contains only R₃. Both branches are connected in parallel across the source.
Equivalent Resistance
- Branch A: ( R_A = 8 Ω + 12 Ω = 20 Ω ).
- Branch B: ( R_B = 24 Ω ).
- Parallel combination:
[ \frac{1}{R_{\text{eq}}} = \frac{1}{20 Ω} + \frac{1}{24 Ω} = \frac{12+10}{240} = \frac{22}{240} = \frac{11}{120} ] Which means, ( R_{\text{eq}} = \frac{120}{11} Ω ≈ 10.91 Ω ).
Total Current
( I_{\text{total}} = 24 V / 10.91 Ω ≈ 2.20 A ) The details matter here..
Current Through R₂
First, find the current in Branch A:
[ I_A = 24 V / 20 Ω = 1.20 A ]
Since R₁ and R₂ are in series, the same current flows through R₂. Because of this, the current through R₂ is 1.20 A.
Circuit 3 – Wheatstone Bridge
Description
- Resistors: R₁ = 10 Ω, R₂ = 20 Ω, R₃ = 30 Ω, R₄ = 60 Ω
- Voltage source: V = 30 V
- Configuration: Classic Wheatstone bridge with a mid‑node connection (no detector). The source is applied across the left–right ends, and the right–left ends are connected through R₃ and R₄.
Equivalent Resistance
The bridge is balanced because:
[ \frac{R_1}{R_2} = \frac{R_3}{R_4} \quad \Rightarrow \quad \frac{10}{20} = \frac{30}{60} = 0.5 ]
When balanced, the mid‑node is at the same potential as the node between R₃ and R₄, so the bridge behaves like two series pairs in parallel:
- Left pair: R₁ + R₂ = 10 Ω + 20 Ω = 30 Ω
- Right pair: R₃ + R₄ = 30 Ω + 60 Ω = 90 Ω
Now, these two series pairs are in parallel:
[ \frac{1}{R_{\text{eq}}} = \frac{1}{30 Ω} + \frac{1}{90 Ω} = \frac{3+1}{90} = \frac{4}{90} = \frac{2}{45} ] [ R_{\text{eq}} = \frac{45}{2} Ω = 22.5 Ω ]
Total Current
( I_{\text{total}} = 30 V / 22.Consider this: 5 Ω = 1. 33 A ).
Voltage Across R₃
Because the bridge is balanced, the voltage across R₃ equals the voltage across R₁ (both are 10 Ω and 30 V applied across the left side). Using the voltage divider:
[ V_{R_3} = I_{\text{total}} \times \frac{R_3}{R_3 + R_4} = 1.33 A \times \frac{30 Ω}{90 Ω} ≈ 0.44 V ]
Circuit 4 – Mixed Series‑Parallel with a Current Source
Description
- Resistors: R₁ = 25 Ω, R₂ = 25 Ω, R₃ = 50 Ω
- Current source: I = 0.4 A (emitter‑directed, pushes current into the network)
- Configuration: R₁ and R₂ are in parallel; this parallel pair is in series with R₃.
Equivalent Resistance of Parallel Pair
[ R_{12} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{25 Ω \times 25 Ω}{50 Ω} = 12.5 Ω ]
Total Resistance
( R_{\text{eq}} = R_{12} + R_3 = 12.On top of that, 5 Ω + 50 Ω = 62. 5 Ω ) And that's really what it comes down to..
Voltage Across the Entire Network
Using the current source, the voltage across the network is:
[ V = I \times R_{\text{eq}} = 0.4 A \times 62.5 Ω = 25 V ]
Current Through R₁
First, find the voltage across the parallel pair:
[ V_{12} = I \times R_{12} = 0.4 A \times 12.5 Ω = 5 V ]
Since R₁ and R₂ share this voltage, the current through R₁ is:
[ I_{R1} = \frac{V_{12}}{R_1} = \frac{5 V}{25 Ω} = 0.20 A ]
Scientific Explanation Behind the Numbers
- Ohm’s Law (V = IR) is the backbone of every calculation. Once the equivalent resistance is known, the overall voltage or current can be found immediately.
- Kirchhoff’s Laws guarantee that the sum of voltage drops around any closed loop equals the applied voltage, and the sum of currents entering a node equals the sum leaving. These principles are implicitly used when reducing series and parallel networks.
- Series vs. Parallel: In series, resistances add directly because the same current flows through each element. In parallel, the reciprocal of the total resistance equals the sum of reciprocals of individual resistances because the same voltage is applied across each branch.
- Wheatstone Bridge Balance: When the ratio condition holds, the bridge’s mid‑node is at the same potential on both sides, effectively shorting the bridge and simplifying the network to two series pairs in parallel.
FAQ
| Question | Answer |
|---|---|
| What if the bridge is unbalanced? | You must solve for node potentials using Kirchhoff’s laws or mesh analysis. Now, the mid‑node will have a non‑zero voltage, and the equivalent resistance changes. |
| *Can I use nodal analysis for these problems?Now, * | Yes, nodal analysis is a powerful alternative, especially for more complex networks. Consider this: it often yields the same results more systematically. |
| Why does the current through R₂ in Circuit 2 equal the current in Branch A? | Because R₁ and R₂ are in series; current is the same through all series components. Which means |
| *What if the current source in Circuit 4 were current‑limited? * | The voltage across the network would be limited by the source’s maximum voltage capability, potentially altering the current distribution. |
It sounds simple, but the gap is usually here.
Conclusion
Activity 1.2 serves as a microcosm of circuit analysis: by mastering series‑parallel reductions, applying Ohm’s law, and understanding the behavior of balanced bridges, students gain confidence in tackling more involved electrical networks. The four example circuits illustrate that, regardless of complexity, a systematic approach—identify series/parallel relationships, compute equivalent resistance, and apply fundamental laws—yields clear, accurate results. Keep practicing with varied resistor values and configurations, and soon these calculations will become second nature.
