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Direct-current electric circuits

An electric circuit is a closed path along which charge flows: a source pushes the current and the elements (resistors, capacitors, inductors, diodes) share it out. Solving it means finding the voltage and current of each element. With two laws (Ohm's and Kirchhoff's) and a few techniques to save work, any direct-current circuit can be solved: that is what the Electric circuits tool does, step by step.

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Voltage, current, Ohm's law and power

Voltage $V$ (in volts) is the potential difference between two points, the «pressure» that moves the charges; current $I$ (in amperes) is the charge that passes per second. In a resistor the two are proportional:

V = R I P = V I = I^{2} R = \frac{V ^{2}}{R}

Power $P$ is energy per second: a source delivers it when the current leaves through its positive terminal, and a resistor always consumes it (it heats up). In a correctly solved circuit, what is delivered and what is consumed match.

Element	Relation	In direct current, at steady state
Resistor	$V = R I$	consumes $I^{2} R$
Capacitor	$I = C \frac{d V}{d t}$ , $Q = C V$	open circuit (no current); stores $\frac{1}{2} C V^{2}$
Inductor	$V = L \frac{d I}{d t}$	short circuit (0 V); stores $\frac{1}{2} L I^{2}$
Real battery	$V = ε - r I$	the terminal voltage drops with the current

Kirchhoff's laws

Current law (nodes): at each node, the sum of the currents entering equals the sum of those leaving, $\sum I = 0$ . It is conservation of charge.
Voltage law (loops): going around any closed path, the sum of the voltage rises and drops is zero, $\sum V = 0$ . It is conservation of energy.

A node is everything joined by wires with no element in between: even if the drawing shows several points, if a wire joins them they are the same node. The simulator names them a, b, c… in its first step.

Series and parallel

Two elements are in series if between them there is a node that nothing else reaches: the same current flows through them. They are in parallel if they go between the same two nodes: they have the same voltage.

	Series	Parallel
Resistors	$R_{e q} = R_{1} + R_{2} + \dots$	$\frac{1}{R _{e q}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \dots$ (dos: $\frac{R _{1} R _{2}}{R _{1} + R _{2}}$ ),(two: $\frac{R _{1} R _{2}}{R _{1} + R _{2}}$ )
Capacitors	$\frac{1}{C _{e q}} = \frac{1}{C _{1}} + \frac{1}{C _{2}} + \dots$	$C_{e q} = C_{1} + C_{2} + \dots$
Divider	voltage: $V_{1} = V \frac{R _{1}}{R _{1} + R _{2}}$	current: $I_{1} = I \frac{R _{2}}{R _{1} + R _{2}}$

The method is to reduce the circuit, step by step, to a single equivalent resistance, find the total current with Ohm's law and then undo the reduction, sharing out voltages (in series) and currents (in parallel).

When there is no series or parallel: Δ–Y

In a bridge (five resistors in a diamond with one diagonal) no pair is in series or in parallel. It is solved by replacing three resistors in a triangle (Δ) with three in a star (Y) that behave the same as seen from the three nodes:

A triangle and its equivalent star between the same nodes a, b and c.

R_{a} = \frac{R _{ab} R _{a c}}{R _{ab} + R _{b c} + R _{a c}} R_{b} = \frac{R _{ab} R _{b c}}{R _{ab} + R _{b c} + R _{a c}} R_{c} = \frac{R _{a c} R _{b c}}{R _{ab} + R _{b c} + R _{a c}}

Each star resistor is the product of the two triangle resistors that reach its node, over the sum of all three. If the bridge is balanced (

R_{1} R_{4} = R_{2} R_{3}

), no current flows through the diagonal and it can be removed.

Several sources: superposition and nodal analysis

With more than one source you can no longer reduce everything to one equivalent. Two ways:

Superposition (linear circuits): leave one source on and turn the others off, a battery is replaced by a wire and a current source by an open circuit. Solve each case and add the results.
Nodal analysis: pick a reference node (0 V), write the current law at each node with the currents written in terms of their voltages, $I = (V_{a} - V_{b}) / R$ , and solve the system. It is the method simulation programs use.

Thévenin equivalent and RC and RL transients

Seen from two terminals, any linear circuit behaves like a source $V_{t h}$ in series with a resistance $R_{t h}$ : $V_{t h}$ is the open-circuit voltage between the terminals and $R_{t h}$ the resistance between them with the sources turned off. It is the quick way to study a capacitor or an inductor connected to a network.

x (t) = x (\infty) + [x (0) - x (\infty)] e^{- t / τ} τ_{RC} = R_{t h} C τ_{R L} = \frac{L}{R _{t h}}

$x$ is the capacitor voltage or the inductor current, the quantities that cannot jump. At $t = τ$ it has covered 63.2 % of the way, and after about $5 τ$ it is considered done (99.3 %).

Charging of the capacitor in example 4.

Common mistakes

Calling two consecutive elements «in series» when a third branch reaches the node between them.
Turning off a current source with a wire (it is turned off with an open circuit) or a battery with an open circuit (it is turned off with a wire).
Adding powers in superposition: only currents and voltages add; the power is computed at the end.
Using the total resistance of the circuit as a capacitor's $R_{t h}$ : it is the one the capacitor sees from its terminals, with the sources off.
Losing the prefixes: 1 kΩ · 100 µF = 0.1 s, not 100 s.

Worked examples

Example 1Series–parallel

A 24 V battery feeds

R_{1} = 4

Ω in series with a parallel made of

R_{2} = 12

Ω and the branch

R_{3} + R_{4}

(

2 + 4

Ω). Find the equivalent resistance, the current and voltage of each resistor and the power.

Solution

$R_{3}$ and $R_{4}$ in series: $6$ Ω. In parallel with $R_{2}$ : $\frac{12 \cdot 6}{12 + 6} = 4$ Ω. In series with $R_{1}$ : $R_{e q} = 8$ Ω.
$I = 24/8 = 3$ A, all of it through $R_{1}$ : $V_{1} = 12$ V. That leaves $24 - 12 = 12$ V for the parallel.
$I_{2} = 12/12 = 1$ A and the branch $R_{3} + R_{4}$ carries $12/6 = 2$ A: $V_{3} = 4$ V, $V_{4} = 8$ V. Check at node b: $3 = 1 + 2$ ✓.
Power: the battery delivers $24 \cdot 3 = 72$ W; the resistors consume $36 + 12 + 8 + 16 = 72$ W ✓.

R_{e q} = 8

Ω,

I = 3

V_{1} = 12

V_{2} = 12

V_{3} = 4

and

V_{4} = 8

P = 72

Solve it in Simuveliq

In Engineering → Electric circuits, in Examples, «Load circuit» → Series–parallel (classic).
The cards give Equivalent resistance 8 Ω, Total current 3 A and Power 72 W. The «Step-by-step reduction» is stepped through with ▶ and shows each series and parallel drawn; below, «Back to the original circuit» shares out voltages and currents.
The Values in each element table has V, I and P of each, with the power balance.

Open in the simulator →

Example 2Wheatstone bridge with Δ–Y

A 10 V source feeds a bridge with

R_{1} = 10

R_{2} = 20

R_{3} = 30

R_{4} = 40

Ω and the diagonal

R_{5} = 50

Ω. Find the equivalent resistance and the current in the diagonal.

Solution

It is not balanced: $R_{1} R_{4} = 400 \neq = R_{2} R_{3} = 600$ . No pair is in series or in parallel.
The triangle $R_{1}, R_{5}, R_{3}$ (nodes a, b, c; sum 90 Ω) becomes a star: $\frac{10 \cdot 30}{90} = 3.333$ Ω toward a, $\frac{10 \cdot 50}{90} = 5.556$ Ω toward b and $\frac{50 \cdot 30}{90} = 16.67$ Ω toward c.
Now it works: $20 + 5.556 = 25.56$ Ω and $40 + 16.67 = 56.67$ Ω in parallel give $17.61$ Ω; plus $3.333$ : $R_{e q} = 20.95$ Ω and $I = 10/20.95 = 477.4$ mA.
Back again: $V_{b} = 6.581$ V and $V_{c} = 5.935$ V (relative to d), so the diagonal carries $\frac{6.581 - 5.935}{50} = 12.9$ mA from b to c.

R_{e q} = 20.95

Ω,

I = 477.4

mA and

I_{5} = 12.9

mA.

Solve it in Simuveliq

«Load circuit» → Wheatstone bridge (Δ–Y). Step 1 of the reduction is «Delta → star (Δ → Y)», with the three formulas; the Equivalent resistance card gives 20.95 Ω.
The «Back from Y to Δ» step finds the actual currents from the node potentials. Change $R_{5}$ or $R_{4}$ until $R_{1} R_{4} = R_{2} R_{3}$ and watch the diagonal current vanish.

Open in the simulator →

Example 3Two batteries: nodal and superposition

V_{1} = 12

V with

R_{1} = 4

Ω and

V_{2} = 6

V with

R_{2} = 2

Ω feed node a, tied to ground through

R_{3} = 6

Ω (both batteries have their positive terminals up). Find the voltage of node a and the current of each battery.

Solution

Nodal, with the bottom at 0 V: $\frac{12 - V _{a}}{4} = \frac{V _{a}}{6} + \frac{V _{a} - 6}{2}$ . Multiplying by 12: $36 - 3 V_{a} = 2 V_{a} + 6 V_{a} - 36$ , so $V_{a} = 72/11 = 6.545$ V.
$I_{1} = (12 - 6.545) /4 = 1.364$ A (it leaves $V_{1}$ : it delivers 16.36 W). Through $R_{2}$ , $(6.545 - 6) /2 = 0.2727$ A enter $V_{2}$ 's positive terminal: it gets charged, absorbing 1.636 W.
By superposition: $V_{1}$ alone ( $R_{e q} = 4 + 6 ∥ 2 = 5.5$ Ω) gives 2.182 A through $R_{1}$ ; $V_{2}$ alone ( $R_{e q} = 2 + 4 ∥ 6 = 4.4$ Ω) gives 1.364 A, of which 0.818 A flow through $R_{1}$ the other way. Total: $2.182 - 0.818 = 1.364$ A ✓.

V_{a} = 6.545

V_{1}

delivers 1.364 A and

V_{2}

absorbs 0.2727 A.

Solve it in Simuveliq

«Load circuit» → Two batteries (superposition). In Solution method pick Superposition to see each source separately («Case: only V1», «Case: only V2») or Nodal for the system of equations.
The source cards say that V1 delivers 16.36 W with 1.364 A and that V2 absorbs 1.636 W; the drawing marks 6.55 V on R3.

Open in the simulator →

Example 4Charging a capacitor (RC)

A 10 V battery,

R_{1} = 1

kΩ in series and

R_{2} = 1

kΩ in parallel with an initially uncharged 100 µF capacitor. Find the capacitor's final voltage, the time constant and

v_{C} (t)

Solution

At the end the capacitor is an open circuit: a 1 k and 1 k divider remains, and $v_{C} (\infty) = V_{t h} = 5$ V.
Resistance seen from the capacitor, with the battery replaced by a wire: $R_{1} ∥ R_{2} = 500$ Ω.
$τ = 500 \cdot 100 \times 1 0^{- 6} = 50$ ms, and since it starts uncharged: $v_{C} (t) = 5 (1 - e^{- t /50 ms})$ V. At 50 ms it is 3.16 V; at 250 ms, 4.97 V.
Initial capacitor current: $V_{t h} / R_{t h} = 10$ mA, which decays with the same time constant. Final energy: $\frac{1}{2} \cdot 100 μ F \cdot 5^{2} = 1.25$ mJ.

v_{C} (\infty) = 5

τ = 50

ms and

v_{C} (t) = 5 (1 - e^{- t /50 ms})

Solve it in Simuveliq

«Load circuit» → RC circuit: capacitor charging.
The «Capacitor charging C1 (RC transient)» block draws the curve and writes the four steps: final voltage 5 V, $R_{t h} = 500$ Ω (with its own step-by-step reduction), $τ = 50$ ms and the response. The table gives Q = 500 µC and U = 1.25 mJ.

Open in the simulator →

Direct-current electric circuits

Voltage, current, Ohm's law and power

Kirchhoff's laws

Series and parallel

When there is no series or parallel: Δ–Y

Several sources: superposition and nodal analysis

Thévenin equivalent and RC and RL transients

Common mistakes

Worked examples

Example 1Series–parallel

Solution

Solve it in Simuveliq

Example 2Wheatstone bridge with Δ–Y

Solution

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Example 3Two batteries: nodal and superposition

Solution

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Example 4Charging a capacitor (RC)

Solution

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