Circuits and electrical energy

6 visual steps · 3 worked examples · 5 exercises with solutions

First, let’s see the idea.

Do two series resistors carry the same current or have the same voltage?

Move forward one step at a time. Change the values and observe what happens.

Current travels around the loopSource: 12 VR₁R₂4 Ω8 Ω
Drawing description

The closed loop connects a 12 V source to 4 Ω and 8 Ω series resistors.

Step 1 of 6

Follow the closed path

Follow the drawing from the source's larger plate until you return to the smaller one.

The wire connects the source to both resistors and returns to the source without branching: they are in series. We assume resistance-free wires, an ideal source and an already steady current.

Source
12 V
Resistors
4 Ω and 8 Ω

What if you change this?

3 V24 V
2 Ω20 Ω
2 Ω20 Ω

Before moving on

With U=12 V, R₁=4 Ω and R₂=8 Ω, calculate I and the voltage drops.

Compare with the explanation

I=1 A, U₁=4 V, U₂=8 V. Total power is 12 W.

You decide when to mark this lesson as studied.

Current travels around the loop

Slide the drawing to see the details.

Explore the full explanationThe foundations, principles, worked examples and all the exercises, at your own pace.

What you will understand

  • Relate current to charge flow and voltage to energy per charge.
  • Solve resistive circuits using Ohm’s and Kirchhoff’s laws.
  • Check power and energy balances.

Before you start

You do not need to know everything by heart. The explanation revisits the necessary ideas and shows how they connect. If a foundation is still unclear, you can return to these lessons:

Charge, current and potential difference

Average current is I=ΔQ/Δt: the amount of charge crossing a section per time interval. The ampere corresponds to coulomb per second. In a metallic conductor, electrons drift opposite to conventional current. The latter is defined by the motion positive charges would have. Choosing a reference direction does not claim that we already know the physical direction.

Potential difference relates the change in electric potential energy to charge: ΔU=qΔV. A volt corresponds to joule per coulomb. Current and voltage are different quantities: current is defined through a section, whereas voltage compares two points. Voltage can exist without current in an open branch. The source does not create circulating charge; it uses another form of energy to maintain a potential difference and transfer energy to the circuit.

Resistance and scope of Ohm’s law

In an ohmic resistor under approximately constant conditions, terminal voltage is proportional to current: V=RI. Resistance R is measured in ohms, Ω=V/A. The relationship describes experimental behaviour; it is not a universal law for all components. Diodes, heating filaments and other elements may have a non-linear or temperature-dependent relationship.

For a homogeneous wire, R=ρℓ/A in a uniform model: ρ is resistivity in Ω·m, ℓ length and A cross-sectional area. A longer wire has greater resistance; a larger cross-section allows lower resistance. Ideal connections are treated as having zero resistance, but real wires dissipate energy. The approximation’s validity depends on how much these voltage drops matter compared with those of the components studied. Circuits here are low-voltage models and exercises.

Nodes and conservation of charge

A node joins branches connected by ideal conductors. In a steady state, there is no significant charge accumulation at the node. Therefore, the sum of entering currents equals the sum of leaving currents. This is Kirchhoff’s current law, often written as the algebraic sum of currents equalling zero. Choose a convention, such as positive entering currents, and keep it throughout the equation.

If 0,8 A enters and 0,3 A leaves along one branch, 0,5 A must leave along the remaining branch. Current is not lost in resistors: electrical energy is transformed there, but charge is conserved. In series without branches, current is the same in every component. In parallel, branches share the same two nodes and voltage; currents may differ according to resistance.

Loops, signs and equivalent resistances

In a lumped circuit in steady state, without significant changing magnetic flux through its loops, the sum of potential changes around a closed loop is zero. Crossing a source from its negative to positive terminal adds a rise +V. Crossing a resistor along the reference current includes a drop −RI. Crossing the resistor in the opposite direction gives +RI.

In series, drops add: V=I(R₁+R₂), so equivalent R is their sum. In parallel, I=V/R₁+V/R₂, so 1/equivalent R=1/R₁+1/R₂. The equivalent resistance of positive parallel resistors is smaller than the smallest branch resistance. A negative calculated current does not invalidate the solution: it means its real direction opposes the chosen one. Always check that equations respect the stated directions.

Power, energy and passive convention

Instantaneous electrical power is p=vi when v and i follow the passive convention: reference current enters the terminal marked positive. Then p>0 means the component absorbs energy and p<0 means it supplies energy. In a resistor with R>0, p=Ri²=v²/R≥0. A source powering the circuit has negative power under this convention because current leaves its positive terminal.

For constant direct current, E=PΔt. A watt is joule per second; a kilowatt-hour is energy: 1 kWh=3,6×10⁶ J. Do not write kW/h for energy consumed. In a model without energy accumulation in fields, the powers of all components sum to zero: what sources supply equals what receivers absorb. This balance independently checks the current equations.

Real sources and responsible interpretation

A real source can be modelled by electromotive force ε and internal resistance r in series. When supplying current I, terminal voltage is V=ε−Ir. The chemical power or power from another conversion mechanism is εI; part, I²r, heats the interior and the remainder reaches the external circuit. Electromotive force is measured in volts despite its historical name: it is not force in newtons.

An ideal ammeter is connected in series and has zero resistance; an ideal voltmeter compares two nodes and has infinite resistance. These ideals help analyse how measurement alters a real circuit. Numerical examples use limited low-voltage sources and conceptual diagrams. Mains installations require equipment and qualified personnel; this unit aims to interpret models and balances, without instructions for work on domestic wiring. Start with nodes and directions, move to equations and finish with units and power.

Worked examples, without skipping steps

Follow the reasoning first. Then cover the solution and try to reconstruct it in your own words.

Example 1

Series with a power balance

An ideal 12 V source powers 4 Ω and 8 Ω series resistors. Calculate current, voltages and powers.

  1. Equivalent resistance is 12 Ω; common current is I=12/12=1 A.
  2. The drops are 4 V and 8 V, summing to 12 V. Absorbed powers are I²R: 4 W and 8 W.
  3. Current leaves the source’s positive terminal; under the passive convention, its power is −12 W.
Result

I=1 A; voltages 4 V and 8 V; total absorbed power 12 W.

Check the result

The loop gives +12−4−8=0 and the balance gives −12+4+8=0 W.

Example 2

Parallel and energy

6 Ω and 3 Ω resistors are connected in parallel to an ideal 6 V source. What energy is transferred in 5 min?

  1. Each branch has 6 V. Currents are 6/6=1 A and 6/3=2 A.
  2. The source supplies 3 A, by the node law. Total power is 6·3=18 W.
  3. Convert 5 min to 300 s. E=18·300=5400 J.
Result

Energy 5400 J; equivalent resistance 2 Ω.

Check the result

Branch powers are 6 W and 12 W. The equivalent 2 Ω is smaller than 3 Ω, as expected.

Example 3

Voltage of a real source

A source has ε=9 V and r=1 Ω, connected to an external 8 Ω resistor. Calculate terminal voltage and internal losses.

  1. Total resistance is 9 Ω; current is ε/(R+r)=9/9=1 A.
  2. Terminal voltage is 9−1·1=8 V, also equal to external IR.
  3. The source converts 9 W; its interior dissipates I²r=1 W and the external resistor absorbs 8 W.
Result

Terminal voltage 8 V; internal dissipation 1 W.

Check the result

9 W=1 W+8 W. Terminal voltage is lower than electromotive force while the source supplies current.

Now try it yourself

There is no timer. Write what you know, choose a relationship and justify each step. Hints and solutions are available whenever you need them.

Exercise 1

18 C passes in 6 s. What is the average current?
A hint
Divide charge by the interval.
See the worked solution
I=18/6=3 A. This is an average; it does not prove that current remained constant at every instant.

Exercise 2

An ohmic 20 Ω resistor receives 4 V. Calculate current and power.
A hint
First use I=V/R.
See the worked solution
I=4/20=0,2 A. Power is VI=4·0,2=0,8 W; V²/R=16/20 confirms the result.

Exercise 3

2,4 A enters a node; 0,9 A and I leave. Determine I.
A hint
Write entering currents equal to leaving currents.
See the worked solution
2,4=0,9+I, so I=1,5 A outwards. If this direction had been chosen as entering, the value would be −1,5 A.

Exercise 4

A 50 W appliance operates for 4 h. What energy is used in kWh and joules?
A hint
Convert power to kilowatts or time to seconds.
See the worked solution
50 W=0,05 kW; E=0,05·4=0,2 kWh. In joules, 0,2·3,6×10⁶=720000 J.

Exercise 5

The potential difference across a component is 5 V. The reference direction for current points into the positive terminal, and the reading in that direction is −0.4 A. Does the component absorb or supply energy?
A hint
Apply the passive convention with the signed value.
See the worked solution
p=5·(−0.4)=−2 W. The component supplies 2 W to the circuit. The negative sign means that conventional current flows opposite to the reference arrow and leaves through the positive terminal.

Mistakes worth avoiding

The resistor consumes current.
Charge is conserved; the resistor transforms energy.
Parallel currents are always equal.
Voltages are equal; currents depend on branch resistances.
Negative power necessarily means an error.
Under the passive convention, it means energy supply.

The essentials to take with you

  • Current is charge per time; voltage is energy per charge.
  • Ohm’s law requires appropriate component behaviour.
  • Kirchhoff expresses conservation with consistent signs.
  • Confirm the circuit using a power balance.

If you still cannot explain one of these ideas, return to the corresponding section and work through an example again. Recognising the difficulty is part of studying.

See the idea in motion

Now you can change the values, observe the graph and compare the experiment with the results you calculated.

Try it in the lab

References for further study

Text and examples prepared for this learning path. These sources help check the principles and continue studying:

The organisation by school year is a study proposal. Pedagogical review by teachers remains a stage of the project.