DC Circuits and Fundamental LawsIntroductionDirect-current (DC) circuit analysis is a foundation of electrical and mechatronics engineering. Sensors, embedded controllers, actuators, battery systems, motor drives, and industrial control circuits all depend on the controlled movement of electric charge. Although practical systems may contain switching devices, semiconductor converters, and nonlinear loads, their first analysis is commonly based on ideal sources, resistors, and the fundamental laws of circuit theory. This chapter develops a systematic method for analyzing resistive DC circuits. It introduces voltage, current, resistance, power, Ohm’s law, Kirchhoff’s current law (KCL), Kirchhoff’s voltage law (KVL), source combinations, voltage and current division, and series-parallel reduction. The emphasis is not only on obtaining numerical answers but also on selecting a valid model, defining reference directions, checking results, and interpreting the engineering meaning of the answer. Learning ObjectivesAfter studying this chapter, you should be able to: - Define voltage, current, resistance, power, energy, nodes, branches, and loops.
- Apply Ohm’s law under appropriate assumptions.
- Reduce series, parallel, and series-parallel resistor networks.
- Apply KCL and KVL using consistent reference directions and polarities.
- Combine ideal voltage and current sources correctly.
- Calculate voltage division, current division, and electrical power.
- Check circuit solutions using conservation laws, units, limiting cases, and power balance.
- Analyze a practical DC sensor or control circuit and identify model limitations.
1. Electrical Quantities and Circuit ModelsAn electrical circuit is an interconnected set of components through which electromagnetic energy can be transferred. Circuit theory uses a lumped-parameter model: component dimensions are assumed small enough that voltage and current can be treated as well-defined quantities at terminals. 1.1 Charge and currentElectric charge is measured in coulombs (C). Current is the rate at which charge crosses a reference surface. For time-varying current, i(t) = dq(t)/dt where i(t) is current in amperes (A), q(t) is charge in coulombs, and t is time in seconds. A current of 1 A corresponds to 1 C/s. Conventional current is defined as the motion of positive charge; in metallic conductors, electrons move in the opposite direction. A current arrow establishes the positive reference direction. If a calculation gives a negative value, the actual current flows opposite to the selected arrow; the circuit analysis is not wrong. 1.2 VoltageVoltage is the energy transferred per unit charge between two points. The voltage from point a to point b is vab = dw/dq where v_ab is in volts (V), w is energy in joules (J), and q is charge in coulombs. One volt equals one joule per coulomb. Voltage is always measured between two points. A voltage polarity marked + and − defines the positive reference: v_ab is positive when point a is at a higher potential than point b according to the chosen reference. 1.3 ResistanceResistance describes opposition to current in a resistive element. Its SI unit is the ohm (Ω). An ideal resistor has a linear voltage-current relationship. Resistance is also related to material and geometry by R = ρℓ/A where R is resistance in ohms, ρ is resistivity in Ω·m, ℓ is conductor length in metres, and A is cross-sectional area in square metres. This relation assumes a uniform material and approximately constant temperature. Real resistance changes with temperature, frequency, mechanical stress, and manufacturing tolerance. 1.4 Power and energyUsing the passive sign convention, current enters the terminal marked positive for voltage v. The instantaneous power absorbed by an element is p = vi where p is in watts (W), v is in volts, and i is in amperes. For a resistor, Ohm’s law gives equivalent forms: p = vi = i²R = v²/R These forms apply to a resistor with resistance R and are dimensionally consistent: V·A = W and A²·Ω = W. If p > 0, the element absorbs power; if p < 0 under the selected reference convention, it delivers power. Energy transferred over a time interval is W = ∫ p(t) dt For constant power, W = Pt. Energy is measured in joules. A battery delivers energy, while a resistor normally converts electrical energy into heat. 2. Ohm’s LawFor an ideal linear resistor, v = iR where v is the voltage across the resistor, i is the current entering its positive terminal, and R is constant in ohms. Rearranged forms are i = v/R and R = v/i when the relevant quantities are known and R is nonzero. Ohm’s law is a component law, not a universal law for every electrical device. It is a useful approximation for metallic resistors over a specified operating range. Diodes, transistors, lamps, batteries, and many sensors may have nonlinear or time-dependent voltage-current relationships. Worked Example 1: Basic resistor calculationA 12 V source is connected across a 2.0 kΩ resistor. Determine the current and resistor power. Using Ohm’s law, i = V/R = 12 V/(2000 Ω) = 0.006 A = 6.0 mA The power is P = VI = 12 V × 0.006 A = 0.072 W = 72 mW The resistor must have a power rating greater than 72 mW. In practice, a 0.25 W resistor would generally provide a reasonable margin, subject to temperature and reliability requirements. 3. Circuit Topology and Fundamental LawsA node is a point at which two or more element terminals are electrically connected. A branch is a single element between two nodes. A loop is a closed path through circuit branches. A reference node, often called ground, is assigned zero voltage; this is a measurement reference and need not be connected to earth. 3.1 Kirchhoff’s current lawKCL follows from conservation of electric charge: charge cannot accumulate indefinitely at an ideal node. The algebraic sum of currents at a node is zero. Σ ik = 0 Currents entering may be assigned positive signs and currents leaving negative signs, or the reverse, provided the convention is used consistently. For example, if currents i₁ and i₂ enter a node and i₃ leaves, i₁ + i₂ − i₃ = 0 or equivalently i₃ = i₁ + i₂. KCL applies at every node in a lumped circuit. At very high frequencies, electromagnetic energy stored in parasitic capacitances and inductances must be included in the circuit model; the generalized conservation principle still holds, but a simple resistive model may not. 3.2 Kirchhoff’s voltage lawKVL follows from conservation of energy. The algebraic sum of voltage rises and drops around any closed loop is zero. Σ vk = 0 A traversal direction must be selected. Crossing a source from − to + is a rise; crossing a resistor in the direction of its reference current is a drop of iR under the passive sign convention. For a loop containing a source V_s and resistors R₁ and R₂ carrying current i, −Vs + iR₁ + iR₂ = 0 Thus i = V_s/(R₁ + R₂). KVL is valid for lumped circuits when voltage is consistently defined. In distributed electromagnetic systems, changing magnetic flux can produce a nonconservative electric field, requiring a more general electromagnetic description. 4. Series CircuitsElements are in series when the same current must pass through each element. This occurs when their common connection has no other branch attached. For resistors in series, KCL implies identical current, and KVL gives Req = R₁ + R₂ + ⋯ + Rn The equivalent resistance is greater than every individual positive resistance. The total source voltage is shared among the resistors according to their resistances. Voltage divisionFor n series resistors connected across a source V_s, the voltage across R_k is Vk = Vs × Rk/(R₁ + R₂ + ⋯ + Rn) This formula assumes ideal wiring, resistors in true series, and an unloaded output. If a load is connected across R_k, the load changes the effective resistance and the original division formula is no longer directly valid. Worked Example 2: Series voltage divisionA 24 V source supplies R₁ = 4 kΩ and R₂ = 8 kΩ in series. The current is i = 24 V/(4 kΩ + 8 kΩ) = 2.0 mA The voltage across R₂ is V₂ = iR₂ = 2.0 mA × 8 kΩ = 16 V The voltage across R₁ is 8 V, and 8 V + 16 V = 24 V, confirming KVL. Since R₂ is twice R₁, it receives twice the voltage. 5. Parallel CircuitsElements are in parallel when they connect between the same two nodes. Consequently, every parallel branch has the same voltage. KCL determines how total current divides among the branches. For resistors in parallel, 1/Req = 1/R₁ + 1/R₂ + ⋯ + 1/Rn For two resistors, a convenient form is Req = R₁R₂/(R₁ + R₂) For positive finite resistances, a parallel equivalent is smaller than the smallest branch resistance. Current divisionFor two parallel resistors carrying total current I, I₁ = I × R₂/(R₁ + R₂) and I₂ = I × R₁/(R₁ + R₂) The current in one branch is inversely related to that branch resistance. These expressions assume exactly two branches and a current source or known total current entering the parallel combination. More generally, first find the common voltage V and use I_k = V/R_k. Worked Example 3: Parallel current divisionA total current of 6 A enters two parallel resistors, R₁ = 3 Ω and R₂ = 6 Ω. The equivalent resistance is Req = (3 × 6)/(3 + 6) Ω = 2 Ω The common voltage is V = IR_eq = 6 A × 2 Ω = 12 V. Therefore, I₁ = 12 V/3 Ω = 4 A and I₂ = 12 V/6 Ω = 2 A. KCL is satisfied because 4 A + 2 A = 6 A. 6. Series-Parallel NetworksMany practical circuits consist of combinations that can be reduced step by step. A valid reduction must preserve the terminal behavior of the part being replaced. The recommended procedure is: - Identify unmistakable series or parallel groups.
- Replace one group by its equivalent resistance.
- Redraw the circuit after each reduction.
- Continue until the total equivalent resistance is obtained.
- Find total source current using Ohm’s law.
- Expand the reductions in reverse order to find individual voltages and currents.
- Verify results with KCL, KVL, and power balance.
Worked Example 4: Series-parallel analysisA 20 V source supplies R₁ = 2 Ω in series with a parallel network of R₂ = 6 Ω and R₃ = 3 Ω. First, R₂₃ = (6 × 3)/(6 + 3) Ω = 2 Ω The total resistance is R_total = 2 Ω + 2 Ω = 4 Ω, so the source current is Itotal = 20 V/4 Ω = 5 A The voltage across R₁ is 5 A × 2 Ω = 10 V. Therefore, the parallel network also has 10 V across it. Branch currents are I₂ = 10 V/6 Ω = 1.667 A and I₃ = 10 V/3 Ω = 3.333 A. Their sum is 5.000 A, agreeing with KCL. Power provides a further check. Source power delivered is 20 V × 5 A = 100 W. Resistor powers are 50 W in R₁, approximately 16.67 W in R₂, and approximately 33.33 W in R₃; their total is 100 W. 7. Combining SourcesSource combinations must be interpreted from terminal connections and polarities, not merely from physical appearance. 7.1 Ideal voltage sources in seriesIdeal voltage sources in series combine algebraically according to polarity: Veq = V₁ + V₂ + ⋯ + Vn A source opposing the selected reference direction contributes negatively. Ideal voltage sources cannot generally be connected directly in parallel unless they have identical voltage and polarity; otherwise the ideal model predicts an undefined or unbounded current. Real sources have internal resistance, which limits current. 7.2 Ideal current sources in parallelIdeal current sources in parallel combine algebraically: Ieq = I₁ + I₂ + ⋯ + In Again, opposing arrows contribute negative values. Ideal current sources cannot generally be connected directly in series unless they impose the same current; practical internal characteristics determine what happens in real systems. 7.3 Internal resistance and practical sourcesA practical voltage source is often modeled as an ideal voltage V_s in series with internal resistance R_s. When a load R_L is connected, the load voltage is VL = Vs × RL/(Rs + RL) The terminal voltage decreases as load current increases. A practical current source can be modeled as an ideal current source in parallel with an internal resistance. These models are important for batteries, laboratory supplies, and sensor interfaces. 8. A Systematic KCL and KVL MethodReduction is efficient for simple resistor networks, but KCL and KVL are more general. A reliable analysis method is: - Choose a reference node and label node voltages.
- Assign arbitrary branch-current directions.
- Mark voltage polarities consistently, preferably using the passive sign convention for resistors.
- Write KCL equations at independent nodes, or KVL equations for independent loops.
- Substitute component relations such as v = iR.
- Solve the simultaneous equations.
- Interpret negative values as reversed assumed directions.
- Verify with an independent law or power balance.
Nodal exampleConsider a node at voltage V connected through R₁ = 2 kΩ to a 10 V node and through R₂ = 3 kΩ to ground. KCL at the unknown node gives (V − 10 V)/2 kΩ + V/3 kΩ = 0 Multiplying by 6 kΩ gives 3V − 30 V + 2V = 0, so 5V = 30 V and V = 6 V. The current from the 10 V source toward the node is (10 − 6)/2 kΩ = 2 mA. The current from the node to ground is 6/3 kΩ = 2 mA, confirming KCL. 9. Electrical Power, Ratings, and EfficiencyPower analysis is essential for selecting components. A resistor’s calculated dissipation must be below its rated power, with an appropriate design margin. Temperature, enclosure ventilation, tolerance, and transient overloads must be considered. For a source and load, efficiency is η = Pload/Pinput × 100% where η is efficiency in percent, P_load is useful load power, and P_input is source input power. The formula assumes powers are measured consistently and that the defined input and useful output boundaries are clear. For the practical voltage source model, maximum load power occurs when R_L = R_s. This is the maximum-power-transfer condition, but the resulting efficiency is only 50%, so it is not usually the preferred condition for power systems. Power supplies generally use loads much larger than source resistance to obtain high efficiency and stable voltage. 10. Applied Case: Battery-Powered Sensor InterfaceBackgroundThe following is an illustrative engineering case. A mechatronic position sensor produces a voltage that is read by a controller with a 5 V analog input. The sensor supply is a nominal 12 V battery, and a resistor divider is proposed to generate approximately 5 V for a monitoring input. ProblemChoose divider resistors so that the no-load output is 5 V from a 12 V source. Then assess the effect of a controller input resistance of 10 kΩ. Analysis and approachFor an unloaded divider with upper resistor R₁ and lower resistor R₂, Vout = 12 V × R₂/(R₁ + R₂) Select R₁ = 14 kΩ and R₂ = 10 kΩ. Then Vout = 12 V × 10/(14 + 10) = 5.00 V However, the 10 kΩ controller input is in parallel with R₂. The effective lower resistance becomes R₂,eff = (10 kΩ × 10 kΩ)/(10 kΩ + 10 kΩ) = 5 kΩ The actual output is therefore Vout,loaded = 12 V × 5/(14 + 5) ≈ 3.16 V The divider is substantially loaded. A better design can use a much smaller divider resistance, provided battery current and resistor power are acceptable, or it can use a buffer amplifier with sufficiently high input impedance. Outcome and learning pointsThe initial calculation was mathematically correct but based on an invalid practical assumption: the output was not unloaded. The case demonstrates that circuit topology must include connected instruments and inputs. It also illustrates a design trade-off: large resistors reduce quiescent battery current but increase loading sensitivity and noise susceptibility; small resistors improve stiffness but consume more power. In a safety-related or energy-limited product, these choices should be documented, tested over tolerance and temperature, and checked against applicable electrical safety and electromagnetic-compatibility requirements. 11. Limitations, Measurement, and Best PracticeIdeal circuit analysis omits wire resistance, contact resistance, source impedance, leakage, temperature coefficients, component tolerance, noise, and parasitic capacitance or inductance. These effects become important when currents are high, signals are small, accuracy is demanding, or switching edges are fast. A digital multimeter has finite input resistance and can load a high-resistance divider. An ammeter must be connected in series and has a small but nonzero burden voltage; connecting an ideal ammeter directly across a voltage source approximates a short circuit and may be hazardous. A voltmeter is connected in parallel, but its input impedance should be much larger than the circuit resistance at the measurement point. Good engineering practice includes: - draw and label the circuit before calculating;
- maintain SI units and use prefixes carefully;
- distinguish circuit ground from protective earth;
- check resistor power ratings and source current limits;
- verify polarity before connecting sources;
- use a current-limited supply during laboratory testing;
- compare calculated and measured values while accounting for tolerance;
- never rely on an ideal model for safety-critical decisions without validation.
Modern circuit simulators can solve large networks and perform tolerance sweeps, but simulation does not replace correct topology, realistic models, measurement, or engineering judgment. AI-assisted design tools may help generate candidate circuits, yet every equation, connection, rating, and safety assumption must be independently verified by a competent engineer. Chapter SummaryDC circuit analysis begins with well-defined voltage and current references and the lumped-circuit model. Ohm’s law relates voltage, current, and resistance for an ideal linear resistor. KCL expresses conservation of charge at nodes, while KVL expresses conservation of energy around closed paths. Series resistors share a common current and add directly; parallel resistors share a common voltage and combine through reciprocal conductance. Voltage division follows from series resistance, and current division follows from parallel resistance. Source combinations depend on polarity and connection type. A robust solution reduces simple networks where possible and uses KCL or KVL for general networks. Every result should be checked using units, limiting behavior, KCL, KVL, and power balance. Practical design must include source resistance, load effects, measurement instruments, tolerances, thermal ratings, and safety constraints. These principles provide the basis for analyzing sensor interfaces, actuator circuits, battery systems, and larger mechatronic electrical subsystems. Key TermsVoltage: Energy transferred per unit charge, measured in volts. Current: Rate of charge flow, measured in amperes. Resistance: Opposition to current in a resistive element, measured in ohms. Resistivity: Material property used to relate resistance to length and area. Power: Rate of energy transfer, measured in watts. Energy: Ability to perform work, measured in joules. Node: Electrical connection point shared by circuit elements. Branch: A circuit element between two nodes. Loop: A closed path through circuit branches. Reference node: Node assigned zero voltage for analysis. Ohm’s law: Relation v = iR for an ideal linear resistor. KCL: Kirchhoff’s current law, expressing charge conservation at a node. KVL: Kirchhoff’s voltage law, expressing energy conservation around a loop. Series circuit: Circuit arrangement in which elements carry the same current. Parallel circuit: Circuit arrangement in which elements share the same two nodes and voltage. Voltage division: Distribution of source voltage among series impedances or resistances. Current division: Distribution of total current among parallel branches. Passive sign convention: Convention in which current enters the positive voltage terminal for absorbed power. Internal resistance: Resistance used to model nonideal source behavior. Equivalent resistance: A resistance that has the same terminal behavior as a network under specified conditions. Review Questions5 Short Answer Questions- What is the physical meaning of one ampere?
- State the assumptions under which Ohm’s law applies to an ideal resistor.
- Why do parallel circuit branches have the same voltage?
- State Kirchhoff’s current law and identify the conservation principle behind it.
- Why can a resistor divider’s output change when a measuring instrument is connected?
5 Long Answer Questions- Derive the equivalent resistance of two resistors in parallel using KCL and Ohm’s law.
- Explain the difference between KCL and KVL, including their physical foundations and typical applications.
- Describe a complete procedure for solving a series-parallel resistor network and explain how the answer should be verified.
- Explain ideal and practical voltage sources, including the role of internal resistance and load resistance.
- Discuss the limitations of ideal DC circuit models in a real mechatronic system.
3 Analytical / Application Questions- A 30 V source feeds a 5 Ω resistor in series with a parallel combination of 10 Ω and 15 Ω. Determine the total current, voltage across the parallel network, current in each branch, and power absorbed by every resistor.
- A 9 V battery has internal resistance 1 Ω and supplies a 4 Ω load. Determine load current, terminal voltage, load power, and source power. Comment on the difference between load power and total power.
- Design a two-resistor divider that produces 3.3 V from a 12 V source when connected to a 33 kΩ input. Explain how you would select resistor values, calculate the loaded output, estimate divider current, and verify power ratings.
Multiple Choice Questions with AnswersQuestion 1A 12 V source is connected to a 3 kΩ resistor. What current flows through the resistor? A. 0.25 mA B. 4 mA C. 36 mA D. 250 mA Correct Answer: B. 4 mA Question 2Two resistors of 4 Ω and 6 Ω are connected in series. What is their equivalent resistance? A. 2.4 Ω B. 5 Ω C. 10 Ω D. 24 Ω Correct Answer: C. 10 Ω Question 3Two resistors of 6 Ω and 3 Ω are connected in parallel. What is their equivalent resistance? A. 2 Ω B. 3 Ω C. 4.5 Ω D. 9 Ω Correct Answer: A. 2 Ω Question 4Which law states that the algebraic sum of currents at a node is zero? A. Ohm’s law B. Kirchhoff’s current law C. Kirchhoff’s voltage law D. Faraday’s law Correct Answer: B. Kirchhoff’s current law Question 5A 20 V source is connected to 2 kΩ and 8 kΩ resistors in series. What voltage appears across the 8 kΩ resistor? A. 2 V B. 4 V C. 16 V D. 18 V Correct Answer: C. 16 V Question 6A total current of 9 A enters parallel resistors of 3 Ω and 6 Ω. What current flows through the 3 Ω resistor? A. 3 A B. 4.5 A C. 6 A D. 9 A Correct Answer: C. 6 A Question 7A resistor has 10 V across it and carries 2 A. What power does it absorb? A. 5 W B. 12 W C. 20 W D. 100 W Correct Answer: C. 20 W Question 8Which connection is generally invalid for unequal ideal voltage sources? A. Equal voltage sources in series with aiding polarity B. Unequal voltage sources directly in parallel C. A voltage source in series with a resistor D. A voltage source connected across a load resistor Correct Answer: B. Unequal voltage sources directly in parallel Question 9A resistor carries 3 A and has resistance 4 Ω. What is its power dissipation? A. 7 W B. 12 W C. 24 W D. 36 W Correct Answer: D. 36 W Question 10In a practical voltage-source model, increasing load current generally causes the terminal voltage to do what? A. Increase without limit B. Remain exactly constant C. Decrease because of internal resistance D. Become equal to zero in every case Correct Answer: C. Decrease because of internal resistance
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