Electric Circuits: free practice, theory and problems

1. Direct current

What is it about?

A DC circuit has voltage and current that do not change direction: batteries, solar panels and electronics powered from a DC bus. Even though AC dominates the power grid, DC analysis is the foundation of all circuit calculations – the same laws (Ohm's law, Kirchhoff's laws) also apply to instantaneous values in AC circuits.

The goal is always the same: find the current and voltage in every branch of the circuit from the sources and resistors.

Concepts and formulas

• Ohm's law: $U = RI$, so $I = U/R$ and $R = U/I$.
• Power in a resistor: $P = UI = RI^2 = U^2/R$.
• Resistors in series: $R_{tot} = R_1 + R_2 + \dots$ (the current is the same in all of them).
• Resistors in parallel: $\dfrac{1}{R_{tot}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots$ (the voltage is the same across all of them).
• Conductance $G = 1/R$, measured in siemens (S). In parallel the conductances add: $G_{tot} = G_1 + G_2 + \dots$
• Voltage divider (two resistors in series across $U$): $U_2 = U\dfrac{R_2}{R_1+R_2}$.
• Current divider (two resistors in parallel with total current $I$): $I_1 = I\dfrac{R_2}{R_1+R_2}$ (most of the current flows through the smaller resistor).
• Kirchhoff's current law (KCL): the sum of currents into a node equals the sum out.
• Kirchhoff's voltage law (KVL): the sum of voltages around a closed loop is zero.

How to solve the problems

1. Draw the circuit and mark the known voltages, currents and resistors.
2. Combine resistors that are pure series or parallel connections, one step at a time, until the circuit is simple enough.
3. Use Ohm's law on the simplest form of the circuit to find the total current from the source.
4. Work backward through the circuit: use the voltage/current divider or KVL/KCL to find the voltage and current in each branch.
5. Check with a power balance or an equation you have not used.

Example

A voltage source of 20 V drives $R_1 = 4$ Ω in series with a parallel combination of $R_2 = 12$ Ω and $R_3 = 6$ Ω. Find the current through $R_3$.

1. Combine the parallel pair: $R_{23} = \dfrac{12\cdot 6}{12+6} = 4$ Ω.
2. Total resistance: $R_{tot} = 4 + 4 = 8$ Ω.
3. Total current: $I = U/R_{tot} = 20/8 = 2.5$ A.
4. Voltage across the parallel pair: $U_{23} = I\cdot R_{23} = 2.5\cdot 4 = 10$ V.
5. Current in $R_3$: $I_3 = U_{23}/R_3 = 10/6 \approx 1.67$ A.

Answer: $I_3 \approx 1.67$ A.

Common mistakes

• Combining resistors that are neither pure series nor pure parallel connections, without simplifying step by step first.
• Using the current divider backward: most of the current flows through the *smallest* resistor, not the largest.
• Mixing volts, milliamps and kilohms without converting to the same base unit (V, A, Ω).
• Forgetting that the voltage is the same across every branch of a parallel combination, not summed.

Concepts in this part

Practise direct current in the app →

2. Capacitors and inductors

What is it about?

Capacitors and inductors are the two fundamental energy-storing components in electronic circuits. A capacitor stores energy in an electric field and "likes" constant voltage, while an inductor stores energy in a magnetic field and "likes" constant current. This resistance to change is the key to everything from time delays and filters to switched-mode power supplies and motor drives.

Because they respond to *change* rather than the instantaneous value, they cannot be described by Ohm's law alone – they need a time constant that tells you how fast the voltage or current reaches its final value.

Concepts and formulas

• Capacitor: $i = C\,\dfrac{du}{dt}$. Charge $Q = CU$. Energy $E = \tfrac12CU^2$.
• Inductor: $u = L\,\dfrac{di}{dt}$. Energy $E = \tfrac12LI^2$.
• Capacitors in parallel: $C_{tot} = C_1 + C_2 + \dots$ (like resistors in series).
• Capacitors in series: $\dfrac{1}{C_{tot}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dots$ (the opposite of resistors – remember that capacitors "flip" the combination rules).
• Inductors in series: $L_{tot} = L_1 + L_2 + \dots$ Inductors in parallel: $\dfrac{1}{L_{tot}} = \dfrac{1}{L_1} + \dfrac{1}{L_2} + \dots$ (the same rule as for resistors).
• Time constant: $\tau = RC$ for a capacitor, $\tau = L/R$ for an inductor.
• Charging from 0 toward $U$: $u(t) = U(1 - e^{-t/\tau})$. Discharging from $U_0$: $u(t) = U_0e^{-t/\tau}$.
• Inductor current building up toward $U/R$: $i(t) = \dfrac{U}{R}(1 - e^{-t/\tau})$.
• In DC steady state: the capacitor is an open circuit ($i=0$), the inductor is a short circuit ($u=0$).

How to solve the problems

1. Find the time constant $\tau$ (RC or $L/R$) and the final value the quantity is heading toward.
2. Set up the correct exponential formula depending on whether it is charging/building up (toward a nonzero final value) or discharging/decaying (toward zero).
3. Plug in the numbers, making sure $t$ and $\tau$ use the same unit.
4. To find the *time* to reach a given value, solve the formula for $t$ by taking the logarithm on both sides.
5. Check the limiting cases: at $t=0$ you should get the starting value, and for large $t$ you should approach the final value.

Example

A capacitor is charged from 0 V toward 10 V through $R = 2$ kΩ, with $C = 100$ µF. How long does it take for the voltage to reach 8 V?

1. Time constant: $\tau = RC = 2000\cdot 100\cdot10^{-6} = 0.2$ s.
2. Charging formula: $u(t) = 10(1 - e^{-t/0.2})$.
3. Set $u = 8$: $8 = 10(1 - e^{-t/0.2})$, so $e^{-t/0.2} = 0.2$.
4. Take the logarithm: $-t/0.2 = \ln 0.2 \approx -1.609$, giving $t \approx 0.322$ s.

Answer: about 0.32 s.

Common mistakes

• Using the charging formula when the component is actually discharging (or the other way around).
• Mixing up the series/parallel rules for capacitors and inductors – they are opposite of each other, and opposite of resistors for the capacitor.
• Forgetting that inductor current (not voltage) is the quantity that cannot jump instantaneously, and that capacitor voltage (not current) cannot jump instantaneously.
• Mixing milli-, micro- and nanofarads/henries without converting to the same base unit.

Concepts in this part

Practise capacitors and inductors in the app →

3. Alternating current

What is it about?

In alternating current (AC), voltage and current reverse direction in step with a sine wave, instead of staying constant as in direct current. This is the shape of the voltage in the power grid, and it makes capacitors and inductors behave very differently than in a DC circuit: they get an "opposition" to current that depends on frequency, called reactance.

The goal of this unit is to be able to compute impedance, current and power in an AC circuit, and to understand the difference between active, reactive and apparent power – something that is essential for everything from motor sizing to the bill from the power company.

Concepts and formulas

• Sinusoidal quantity: $u(t) = U_p\sin(\omega t + \varphi)$, with peak value $U_p$, angular frequency $\omega = 2\pi f$ rad/s and period $T = 1/f$.
• RMS value: $U = U_p/\sqrt2$ for a pure sine wave. The 230 V mains voltage is always an RMS value.
• Reactance: inductor $X_L = \omega L = 2\pi fL$, capacitor $X_C = \dfrac{1}{\omega C} = \dfrac{1}{2\pi fC}$.
• Impedance: $Z = R + jX$, with magnitude $|Z| = \sqrt{R^2+X^2}$ and phase angle $\varphi = \arctan(X/R)$.
• Current: $I = U/|Z|$, phase-shifted by $\varphi$ relative to the voltage.
• Power: active power $P = UI\cos\varphi$ (W), reactive power $Q = UI\sin\varphi$ (VAr), apparent power $S = UI$ (VA), with $S^2 = P^2+Q^2$ and power factor $\cos\varphi = P/S$.
• Resonance in a series RLC circuit: $X_L = X_C$ at $f_0 = \dfrac{1}{2\pi\sqrt{LC}}$; then $Z=R$ (purely resistive).
• Three-phase (balanced star connection): line voltage $U_L = \sqrt3\,U_f$; total active power $P = \sqrt3\,U_LI_L\cos\varphi$.

How to solve the problems

1. Compute the angular frequency $\omega = 2\pi f$ if you need it.
2. Find the reactance(s) of the inductors/capacitors at the given frequency.
3. Set up the impedance $Z=R+jX$ (inductive reactance is positive, capacitive is negative) and find $|Z|$ and $\varphi$.
4. Use $I=U/|Z|$ for the current, and $P=UI\cos\varphi$, $Q=UI\sin\varphi$, $S=UI$ for the powers.
5. For three-phase problems, check whether the given voltage is a line or a phase voltage, and apply the $\sqrt3$ factor the right way.

Example

A series circuit with $R=30$ Ω and an inductor $L=0.1$ H is connected to a 230 V, 50 Hz source. Find the current and the power factor.

1. Reactance: $X_L = 2\pi fL = 2\pi\cdot50\cdot0.1\approx31.4$ Ω.
2. Impedance: $|Z|=\sqrt{30^2+31.4^2}=\sqrt{900+986}\approx43.4$ Ω.
3. Current: $I=U/|Z|=230/43.4\approx5.29$ A.
4. Phase angle: $\varphi=\arctan(31.4/30)\approx46.3^\circ$, so $\cos\varphi\approx0.69$.

Answer: $I\approx5.29$ A, $\cos\varphi\approx0.69$ (inductive, the current lags the voltage).

Common mistakes

• Mixing up peak value and RMS value – the 230 V mains voltage is always an RMS value.
• Forgetting that $X_L$ and $X_C$ have opposite signs in the impedance, and that $X_C$ *decreases* as frequency increases (the opposite of $X_L$).
• Using the phase voltage where the line voltage is given, or the other way around, in three-phase calculations.
• Thinking that apparent power $S$ can be used directly as "useful" power – only the active power $P$ does useful work.

Concepts in this part

Practise alternating current in the app →

4. Network analysis

What is it about?

Network analysis is about systematic methods for finding the current and voltage in circuits that are too complicated to simplify with series/parallel combinations alone. Instead of starting from scratch every time, engineers use standard recipes: the node-voltage method, the mesh-current method, superposition, and Thévenin/Norton equivalents.

These methods are especially useful when a circuit must be analyzed for many different loads, or when you only care about what happens in one particular branch of a large circuit.

Concepts and formulas

• Node-voltage method: choose one node as ground, and write KCL for the remaining $n-1$ nodes with the node voltages as unknowns.
• Mesh (loop) method: write KVL for each independent loop with mesh currents as unknowns. A planar circuit with $b$ branches and $n$ nodes has $m=b-n+1$ independent loops.
• Superposition: in a linear circuit with several independent sources, the response (current or voltage) is the sum of the contributions from each source acting alone, with the other sources turned off (voltage sources shorted, current sources opened).
• Thévenin equivalent: any linear two-terminal circuit can be replaced by a single voltage source $U_{th}$ in series with a single resistor $R_{th}$, as seen from two given terminals.
• Norton equivalent: the same circuit as a current source $I_N = U_{th}/R_{th}$ in parallel with $R_{th}$.
• $R_{th}$ is found by turning off all independent sources and computing the resistance seen from the terminals, or from $R_{th} = U_{open}/I_{short}$.
• Maximum power transfer to a load occurs when $R_L = R_{th}$, giving $P_{max} = U_{th}^2/(4R_{th})$.

How to solve the problems

1. Choose the method based on what you need: superposition or Thévenin when you only want to know what happens in one branch, the node/mesh method for the whole circuit.
2. With superposition: turn off all sources except one, compute its contribution, repeat for each source, and add the results at the end (never for power).
3. With Thévenin: find $U_{th}$ as the open-circuit voltage across the terminals, and $R_{th}$ with the sources turned off.
4. Put the unknown load back onto the simplified (Thévenin/Norton) circuit and solve with Ohm's law.
5. Check the answer: do the limiting cases ($R_L=0$ or $R_L\to\infty$) give sensible currents and voltages?

Example

A circuit has a voltage source $U_1 = 12$ V in series with $R_1 = 4$ Ω feeding a node, and a current source $I_2 = 2$ A injecting current into the same node. A load resistor $R_2 = 6$ Ω runs from the node to ground (the return path for both sources). Find the current through $R_2$ using superposition.

1. Turn off the current source (open it): $R_1$ and $R_2$ are then in series across $U_1$. Contribution: $I' = U_1/(R_1+R_2) = 12/10 = 1.2$ A.
2. Turn off the voltage source (short it): $R_1$ is now in parallel with $R_2$, seen from the current source. The current splits: $I'' = I_2R_1/(R_1+R_2) = 2\cdot4/10 = 0.8$ A.
3. Add the contributions: $I_{R2} = I' + I'' = 1.2 + 0.8 = 2$ A.

Answer: $I_{R2} = 2$ A.

Common mistakes

• Forgetting to turn off *all* the other sources during superposition (a voltage source is shorted to 0 V, a current source is opened to 0 A, not the other way around).
• Applying superposition to power – power is quadratic, so the contributions cannot simply be added.
• Confusing $R_{th}$ with just "the nearest resistor" – it must be computed from the whole circuit with the sources turned off.
• Setting $R_L = R_{th}$ to maximize *efficiency* – the efficiency is then actually only 50%; maximum power and best efficiency are not the same thing.

Concepts in this part

Practise network analysis in the app →

5. Three-phase and power

What is it about?

The electricity in the wall socket is alternating current, and in AC circuits it is not enough to multiply voltage and current to get the power. Motors and transformers also draw reactive power that flows back and forth without doing useful work. Larger machines are connected to three-phase, which gives smoother power and less copper in the cables.

Concepts and formulas

• Real power $P$ (W) does useful work. Reactive power $Q$ (var) oscillates between source and load. Apparent power $S$ (VA) is what the cables must handle.

• The power factor $\cos\varphi = P/S$ tells how much of the current does useful work. Pure resistors (electric heaters) have $\cos\varphi = 1$.
• Three-phase: three voltages shifted $120^\circ$. The line voltage (between two phases) is $\sqrt3$ times the phase voltage (phase to neutral): $U_L = \sqrt3\,U_f$, for example $\sqrt3\cdot 230 \approx 400$ V.
• Power in a balanced three-phase load:

• Norway has both IT networks with 230 V between phases and TN networks with 230 V phase–neutral and 400 V between phases.

How to solve the problems

1. Find out whether it is single-phase or three-phase, and whether the voltage is a line voltage or a phase voltage.
2. Use $P = UI\cos\varphi$ (single-phase) or $P = \sqrt3\,U_L I_L\cos\varphi$ (three-phase).
3. To find the current, rearrange: $I = P/(\sqrt3\,U_L\cos\varphi)$.

Example

An 11 kW three-phase motor is connected to 400 V and has $\cos\varphi = 0.85$. What is the line current?

1. $I = \dfrac{P}{\sqrt3\,U_L\cos\varphi} = \dfrac{11\,000}{\sqrt3\cdot 400\cdot 0.85}$.
2. $I \approx 18.7$ A.
3. The same power on a single phase at 230 V and $\cos\varphi = 0.85$ would have needed $11\,000/(230\cdot 0.85) \approx 56$ A.

Common mistakes

• Forgetting $\cos\varphi$ and computing $S$ instead of $P$.
• Using the phase voltage in the three-phase formula with $\sqrt3$. The formula uses the line voltage.
• Adding $P$ and $Q$ directly. They are perpendicular: $S = \sqrt{P^2 + Q^2}$.

Concepts in this part

Practise three-phase and power in the app →