Before you can use theorems like Thévenin or Superposition, you need to be fluent in the basic quantities that describe every D.C. circuit: voltage, current, resistance, power and energy — and how resistors, capacitors and inductors each behave once direct current is applied.
📋 The Five Quantities You'll Use in Every D.C. Problem
Voltage (V): the electrical "push" that drives charge around a circuit, measured in volts (V). It is the energy given to each unit of charge that passes through a component.
Current (I): the rate of flow of electric charge, measured in amperes (A). One amp is one coulomb of charge passing a point every second.
Resistance (R): a component's opposition to current flow, measured in ohms (Ω). It converts electrical energy into heat as current passes through it.
Power (P): the rate at which energy is transferred or converted, measured in watts (W). One watt is one joule of energy every second.
Energy (E): the total amount of work done or heat produced over a period of time, measured in joules (J), found by multiplying power by time.
💡
How the Three Components Differ in D.C.
Same voltage source, three very different behaviours
SteadyResistor
Obeys Ohm's Law at every instant. Current and voltage rise and fall together, with no delay — a resistor's behaviour never changes over time in a D.C. circuit.
Charges then blocksCapacitor
Allows current to flow only while it is charging. Once fully charged, it blocks D.C. current completely and behaves like an open circuit.
Resists changeInductor
Opposes any change in current at the moment it happens. Once current is steady, it behaves like a plain wire (a short circuit) with no resistance to D.C.
🧠 Key Words You'll Meet
Direct current (D.C.) — current that flows in one direction only, unlike alternating current (A.C.) which reverses direction periodically.
Steady state — the condition a circuit settles into once any charging or switching transient has finished.
Time constant — a measure of how quickly a capacitor or inductor reaches steady state after a voltage is applied.
Charge (Q) — the quantity of electricity, measured in coulombs (C), that has passed a point or is stored on a capacitor.
Where this fits: everything in this file is the groundwork the circuit theorems build on. Once you're confident with Ohm's Law, power, energy, and how R, C and L each behave, the Thévenin, Norton, Mesh, Superposition and Maximum Power Transfer files will make far more sense.
1
Ohm's Law
The single most important relationship in D.C. circuit theory
Ohm's LawV = I × R
📋 What Each Symbol Means
V = I × R. V = voltage across the component, measured in volts. I = current flowing through the component, measured in amps. R = the component's resistance, measured in ohms.
💡 Where Does V = IR Come From?
Ohm's Law is an experimental relationship: for a resistor at constant temperature, the current flowing through it is directly proportional to the voltage placed across it. Double the voltage, and the current doubles too — the ratio V ÷ I stays constant, and that constant is what we call the resistance, R.
Sanity check: if R = 0 (a plain wire), then for any current to flow at all, V must be 0 — consistent with a wire having no voltage dropped across it. If R is very large (nearly an open circuit), even a large voltage only pushes a tiny current through — which matches how an open switch behaves.
📝Worked Example 1 — Finding Current
GivenA 12 V battery is connected across a 4 Ω resistor.
FindThe current I flowing through the resistor.
Why divide by R? Ohm's Law says V = I × R, so rearranging for I means dividing both sides by R: I = V ÷ R.
WorkingI = V ÷ R = 12 ÷ 4 = 3 A
AnswerI = 3 A
📝Worked Example 2 — Finding Resistance
GivenA component carries 0.5 A when 9 V is applied across it.
FindThe resistance R of the component.
Why divide V by I here? Rearranging V = I × R for R means dividing both sides by I: R = V ÷ I.
WorkingR = V ÷ I = 9 ÷ 0.5 = 18 Ω
AnswerR = 18 Ω
🧠 Recap
Ohm's Law — V = IR, linking voltage, current and resistance for a resistor.
Linear component — one where V is directly proportional to I, i.e. its resistance stays constant regardless of voltage.
🔧 Try It Yourself
Now you can try these calculators with your own values.
⚡Ohm's Law Calculator
Fill in any two values and leave the third blank, then press Calculate.
2
Power & Energy in D.C. Networks
How fast energy is used, and how much energy is used in total
PowerP = V × I = I² × R = V² ÷ R
📋 What Each Symbol Means
P = VI. P = power delivered or dissipated, in watts. V = voltage across the component, in volts. I = current through the component, in amps.
E = P × t. E = energy transferred, in joules. P = power, in watts. t = time the power is sustained for, in seconds.
💡 Where Does P = VI Come From?
Voltage is defined as the energy given to each unit of charge (joules per coulomb), and current is the rate at which charge flows (coulombs per second). Multiplying them together: (joules per coulomb) × (coulombs per second) = joules per second, which is exactly the definition of power in watts. So P = VI falls directly out of the definitions of V and I.
Sanity check: if either V or I is zero, no energy can be transferred each second, so P must be zero — which matches P = VI giving zero whenever either factor is zero.
💡 Where Do P = I²R and P = V²/R Come From?
These are just P = VI with Ohm's Law substituted in. Since V = IR, replacing V in P = VI gives P = (IR) × I = I²R. Since I = V/R, replacing I in P = VI gives P = V × (V/R) = V²/R. Both are the same physical power — just written using whichever two quantities you already know.
Sanity check: for a fixed resistor, doubling the current should quadruple the heating power (since I²R depends on current squared) — this matches how a wire carrying too much current heats up dramatically faster than one carrying just a little more.
📝Worked Example 3 — Power from V and I
GivenA 6 V supply drives 2 A through a heating element.
FindThe power P dissipated by the element.
Why use P = VI here? Both V and I are already known, so the direct definition of power applies without needing Ohm's Law first.
WorkingP = V × I = 6 × 2 = 12 W
AnswerP = 12 W
📝Worked Example 4 — Energy Over Time
GivenA 12 W lamp is left on for 5 minutes.
FindThe energy E consumed, in joules.
Why convert minutes to seconds? Energy in joules requires time in seconds, since a watt is defined as one joule per second — mixing units would give a wrong answer.
Workingt = 5 × 60 = 300 s. E = P × t = 12 × 300 = 3600 J
AnswerE = 3600 J
🧠 Recap
Power (P) — the rate of energy transfer; P = VI = I²R = V²/R.
Energy (E) — the total transferred over time; E = Pt, in joules when P is in watts and t is in seconds.
🔧 Try It Yourself
Now you can try these calculators with your own values.
⚡Power Calculator
Enter V and I to find P = V × I.
⏳Energy Calculator
Enter P and t (in seconds) to find E = P × t.
3
Resistors in D.C. Networks
Series, parallel, and how resistance combines
SeriesRtotal = R1 + R2 + R3 + ...
Parallel1/Rtotal = 1/R1 + 1/R2 + 1/R3 + ...
📋 What Each Symbol Means
Series: Rtotal = the single equivalent resistance replacing the chain. R1, R2, R3 = the individual resistors connected end to end, so the same current flows through each one.
Parallel: Rtotal = the single equivalent resistance replacing the group. R1, R2, R3 = the individual resistors connected across the same two nodes, so each has the same voltage across it.
💡 Why Do Series Resistances Simply Add?
In series, the same current I flows through every resistor. By Ohm's Law, the voltage dropped across each one is I × R1, I × R2, and so on. The total voltage across the whole chain is the sum of these individual drops, so Vtotal = I(R1 + R2 + ...). Comparing this with Vtotal = I × Rtotal shows Rtotal must equal the sum of the individual resistances.
Sanity check: adding more resistors in series can only make it harder for current to flow, never easier, so Rtotal should always increase — which matches a simple sum, since every term is positive.
💡 Why Does Parallel Resistance Use Reciprocals?
In parallel, every resistor shares the same voltage V. By Ohm's Law, the current through each branch is V/R1, V/R2, and so on. The total current drawn from the supply is the sum of these branch currents: Itotal = V(1/R1 + 1/R2 + ...). Since Itotal = V/Rtotal, dividing through by V shows 1/Rtotal equals the sum of the reciprocals.
Sanity check: adding another path for current to flow through can only make it easier for current overall, so Rtotal should always decrease as more resistors are added in parallel — which matches the fact that adding positive terms to 1/Rtotal makes Rtotal itself smaller.
📝Worked Example 5 — Series Combination
GivenThree resistors, 2 Ω, 3 Ω and 5 Ω, are connected in series.
FindThe total resistance Rtotal.
Why simply add them? The same current flows through all three, so their individual voltage drops stack up, meaning their resistances add directly.
WorkingRtotal = 2 + 3 + 5 = 10 Ω
AnswerRtotal = 10 Ω
📝Worked Example 6 — Parallel Combination
GivenTwo resistors, 6 Ω and 3 Ω, are connected in parallel.
FindThe total resistance Rtotal.
Why use reciprocals? Both resistors share the same voltage, so their branch currents add, which only works cleanly using 1/R terms.
Series connection — components joined end to end, sharing the same current.
Parallel connection — components joined across the same two nodes, sharing the same voltage.
🔧 Try It Yourself
Now you can try these calculators with your own values.
🔌Series / Parallel Calculator (2 resistors)
Enter two resistor values to see both the series and parallel totals.
4
Capacitors in D.C. Networks
They charge, then block — and they store energy in an electric field
ChargeQ = C × V
Energy StoredE = ½ × C × V²
📋 What Each Symbol Means
Q = CV. Q = charge stored on the capacitor's plates, in coulombs. C = capacitance, a measure of how much charge the capacitor can store per volt, in farads. V = the voltage across the capacitor's plates, in volts.
E = ½CV². E = energy stored in the capacitor's electric field, in joules. C = capacitance, in farads. V = the voltage across the capacitor once charged, in volts.
💡 Why Do Capacitors Block D.C. Once Charged?
A capacitor stores charge by building up opposite charges on two plates separated by an insulator. As it charges, current flows to deposit that charge — but once the plates reach the same voltage as the supply, there is no further voltage difference driving charge across the insulating gap, so current drops to zero. From that point on, the capacitor behaves like an open circuit to steady D.C.
Sanity check: immediately after a D.C. supply is connected to an uncharged capacitor, the capacitor's voltage is 0, so almost the full supply voltage appears across whatever resistance is in series with it — meaning current is highest at the very start and falls to zero as the capacitor finishes charging.
📝Worked Example 7 — Charge Stored
GivenA 200 µF capacitor is charged to 9 V.
FindThe charge Q stored on the capacitor.
Why multiply C by V? Capacitance is defined as charge stored per volt, so rearranging that definition gives Q = C × V directly.
WorkingQ = C × V = (200 × 10-6) × 9 = 1800 × 10-6 C
AnswerQ = 1.8 × 10-3 C (1.8 mC)
🧠 Recap
Capacitance (C) — the ability to store charge per volt applied, measured in farads (F).
Time constant (τ = RC) — how quickly a capacitor charges through a series resistor; a larger τ means slower charging.
🔧 Try It Yourself
Now you can try these calculators with your own values.
🔌Capacitor Charge & Energy Calculator
Enter capacitance (in microfarads) and voltage to find charge and stored energy.
5
Inductors in D.C. Networks
They resist change in current, then act like a wire — storing energy in a magnetic field
Energy StoredE = ½ × L × I²
📋 What Each Symbol Means
E = ½LI². E = energy stored in the inductor's magnetic field, in joules. L = inductance, a measure of how strongly the component opposes changes in current, in henries (H). I = the steady current flowing through the inductor, in amps.
💡 Why Do Inductors Resist Changes in Current?
An inductor is usually a coil of wire, and any change in the current through it changes the magnetic field it creates. By Faraday's and Lenz's laws, a changing magnetic field induces a voltage (a "back-EMF") that opposes the change causing it. This means an inductor fights against sudden increases or decreases in current — but once the current is steady, the magnetic field is no longer changing, so no back-EMF is produced and the inductor behaves like a plain wire (zero resistance) to D.C.
Sanity check: immediately after a D.C. supply is connected to an inductor with no initial current, the current cannot jump instantly to its final value — it must rise gradually, which matches the idea that the inductor initially opposes the change most strongly and then "relaxes" as current approaches its steady value.
📝Worked Example 8 — Energy Stored in an Inductor
GivenA 0.4 H inductor carries a steady current of 5 A.
FindThe energy E stored in its magnetic field.
Why use ½LI² rather than LI? Energy stored builds up gradually as current rises from 0 to its final value, and integrating that build-up gives the one-half factor and the square on current — the same pattern as kinetic energy building up as speed increases.
Inductance (L) — a measure of how strongly a component opposes changes in current, measured in henries (H).
Time constant (τ = L/R) — how quickly current through an inductor reaches its steady value; a larger τ means a slower rise.
🔧 Try It Yourself
Now you can try these calculators with your own values.
🔌Inductor Energy Calculator
Enter L and I to find the energy stored in the magnetic field.
⚙️
Applications
Using these quantities to solve real workplace situations
🛠️ Sizing a Heating Element
A technician is specifying a small immersion heater for a test rig. The heater must dissipate 500 W when connected to a 230 V D.C. supply, and the technician needs to know what resistance the heating element must have so it can be ordered from a supplier.
WorkingRearranging P = V²/R for R: R = V² ÷ P = 230² ÷ 500 = 52900 ÷ 500 = 105.8 Ω
AnswerThe element must have a resistance of approximately 105.8 Ω.
🔋️ Checking a Fuse Rating
A control panel draws current from a 24 V D.C. supply through a circuit with a total resistance of 8 Ω. The panel is protected by a 2 A fuse, and the maintenance engineer needs to check whether the circuit's normal operating current is safely below the fuse rating before the panel is switched on.
WorkingI = V ÷ R = 24 ÷ 8 = 3 A
AnswerThe circuit draws 3 A, which exceeds the 2 A fuse rating — the fuse would blow, so a higher-rated fuse or lower-resistance path is needed before the panel can be switched on safely.
🔋 Predicting a Smoothing Capacitor's Stored Energy
A power-supply designer is choosing a smoothing capacitor rated at 1000 µF for a 15 V D.C. rail. Before ordering, they want to know how much energy the capacitor will store once fully charged, since this affects how long it can hold up the output voltage during a brief supply interruption.
AnswerThe capacitor stores 0.1125 J — enough to briefly sustain the rail during short interruptions, but the designer would need a larger capacitor for longer hold-up times.
⚡ Estimating Running Cost of Equipment
A workshop supervisor wants to estimate the daily running cost of a 12 V D.C. motor that draws 4 A continuously for an 8-hour shift, in order to include it in an energy budget for the site.
WorkingP = V × I = 12 × 4 = 48 W. t = 8 × 3600 = 28800 s. E = P × t = 48 × 28800 = 1382400 J
AnswerThe motor uses 1,382,400 J (about 0.384 kWh) per shift, which the supervisor can multiply by the site's electricity rate to estimate daily cost.
🔧 Try It Yourself
Now you can try these calculators with your own values.
🛠️All-in-One D.C. Calculator
Enter any two of V, I, R to find the third, plus the resulting power.
