LNGR · Circuit workshop

Read circuits. Step by step.

Ten simple circuits, fifty small tasks. Begin with one current loop and work towards filters. Each answer stays hidden until you need it.

How to read these drawings

Lines are wires. Black dots mark electrical junctions. R denotes resistors, C capacitors, L inductors and D diodes. A, B and M are test points, not additional components. B is our 0 V reference, not automatically protective earth or ground.

U_AB means voltage at A relative to B. A voltmeter connects in parallel: red probe at A, black at B. An ammeter instead goes in series in an opened current path; it does not connect directly across a voltage source.

These begin as paper and model exercises with ideal components. For a physical build, use a current-limited low-voltage source and check component values, polarity and ratings. The sine source is a small signal, not mains voltage.

A closed circuit
Two resistors in series
Two parallel current paths
Dividing voltage — with a load
An LED with a series resistor
The diode as a current valve
Charging a capacitor
An inductor builds up current
RC low-pass: measure across C
RC high-pass: measure across R

Circuit 01 · 5 small steps

A closed circuit

The ideal DC source holds A at +5 V relative to B. Wires are ideal. One resistor closes the loop. Conventional current flows externally from A through R1 to B.

Trace the current path. Where does it return?
Hint
The source needs a closed external circuit.
Answer and working
From A through R1 to B and back to the source. Current is not used up in the resistor; electrical energy becomes heat.
What voltage is measured between A and B?
Hint
The probes are connected across the resistor.
Answer and working
Calculation notes
$U_{AB}=+5\,\mathrm{V}$
$U_{BA}=-5\,\mathrm{V}$
Probe polarity determines the sign.
Convert 1 kΩ to Ω and calculate I.
Hint
I = U/R; 1 kΩ = 1000 Ω.
Answer and working
Calculation notes
$R=1\,\mathrm{k\Omega}=1000\,\Omega$
$I=\frac{U}{R}$
$I=\frac{5\,\mathrm{V}}{1000\,\Omega}$
${I=0.005\,\mathrm{A}=5\,\mathrm{mA}}$
How much power does R1 dissipate?
Hint
P = U × I. Use amperes for current.
Answer and working
Calculation notes
$P=U\cdot I$
$P=5\,\mathrm{V}\cdot0.005\,\mathrm{A}$
${P=0.025\,\mathrm{W}=25\,\mathrm{mW}}$
Imagine doubling R1 to 2 kΩ. What happens? Check the model.
Hint
The source keeps U constant.
Answer and working
Calculation notes
$I=\frac{5\,\mathrm{V}}{2000\,\Omega}=2.5\,\mathrm{mA}$
$P=5\,\mathrm{V}\cdot0.0025\,\mathrm{A}$
${P=12.5\,\mathrm{mW}}$
At fixed voltage, doubling resistance halves current and power.

Think it through before opening the answer. Always write the formula, substituted values and units.

Circuit 02 · 5 small steps

Two resistors in series

R1 and R2 are connected in series. M is their junction. No additional load is connected at M. B is the 0 V reference.

Are there one or two paths from A to B?
Hint
Look for a branch.
Answer and working
There is exactly one path. Both resistors carry the same current.
Calculate the total resistance.
Hint
Series resistances add.
Answer and working
Calculation notes
$R_{\mathrm{total}}=R_1+R_2$
$R_{\mathrm{total}}=1000\,\Omega+2000\,\Omega$
${R_{\mathrm{total}}=3000\,\Omega=3\,\mathrm{k\Omega}}$
Find the current through R1 and R2.
Hint
Use the full source voltage and total resistance.
Answer and working
Calculation notes
$I=\frac{U}{R_{\mathrm{total}}}$
$I=\frac{9\,\mathrm{V}}{3000\,\Omega}$
${I_1=I_2=3\,\mathrm{mA}}$
Find U_AM and U_MB.
Hint
For each resistor, U = R × I.
Answer and working
Calculation notes
$U_{AM}=R_1\cdot I$
$U_{AM}=1000\,\Omega\cdot0.003\,\mathrm{A}=3\,\mathrm{V}$
$U_{MB}=2000\,\Omega\cdot0.003\,\mathrm{A}=6\,\mathrm{V}$
${U_{AM}+U_{MB}=3\,\mathrm{V}+6\,\mathrm{V}=9\,\mathrm{V}}$
What do red at M and black at B read? What if R1 and R2 swap places?
Hint
You measure across the lower resistor only.
Answer and working
Calculation notes
$U_{MB}=6\,\mathrm{V}$
$R_1\leftrightarrow R_2$
$U_{MB}=1000\,\Omega\cdot0.003\,\mathrm{A}$
${U_{MB}=3\,\mathrm{V}}$
After swapping, the smaller resistor is below. Total current stays the same.

Think it through before opening the answer. Always write the formula, substituted values and units.

Circuit 03 · 5 small steps

Two parallel current paths

Both resistors connect to the same nodes A and B. Black dots mark connected wires. The source maintains 6 V.

What is the voltage across each resistor?
Hint
Their terminals connect to the same two nodes.
Answer and working
Calculation notes
${U_{R1}=U_{R2}=U_{AB}=6\,\mathrm{V}}$
Voltage does not split here.
Calculate current in the first branch.
Hint
I1 = U/R1.
Answer and working
Calculation notes
$I_1=\frac{U}{R_1}$
$I_1=\frac{6\,\mathrm{V}}{1000\,\Omega}$
${I_1=0.006\,\mathrm{A}=6\,\mathrm{mA}}$
Calculate current in the second branch.
Hint
I2 = U/R2.
Answer and working
Calculation notes
$I_2=\frac{U}{R_2}$
$I_2=\frac{6\,\mathrm{V}}{2000\,\Omega}$
${I_2=0.003\,\mathrm{A}=3\,\mathrm{mA}}$
How much current must the source supply?
Hint
Branch currents add at the node.
Answer and working
Calculation notes
$I_{\mathrm{total}}=I_1+I_2$
$I_{\mathrm{total}}=6\,\mathrm{mA}+3\,\mathrm{mA}=9\,\mathrm{mA}$
$R_{\mathrm{total}}=\frac{U}{I_{\mathrm{total}}}$
${R_{\mathrm{total}}=\frac{6\,\mathrm{V}}{0.009\,\mathrm{A}}\approx667\,\Omega}$
R2 is disconnected. Does current in R1 increase?
Hint
The ideal source still maintains 6 V.
Answer and working
Calculation notes
$I_2=0$
$I_1=\frac{6\,\mathrm{V}}{1000\,\Omega}=6\,\mathrm{mA}$
${I_{\mathrm{total}}=6\,\mathrm{mA}}$
Current in the first branch stays the same. Only total current falls.

Think it through before opening the answer. Always write the formula, substituted values and units.

Circuit 04 · 5 small steps

Dividing voltage — with a load

The two resistors represent a 10 kΩ potentiometer at its midpoint. M is the wiper. Initially, the M–B output is unloaded.

What is the resistance between A and B?
Hint
Add R1 and R2.
Answer and working
Calculation notes
$R_{\mathrm{total}}=R_1+R_2$
${R_{\mathrm{total}}=5\,\mathrm{k\Omega}+5\,\mathrm{k\Omega}=10\,\mathrm{k\Omega}}$
What current flows through the unloaded divider?
Hint
I = 5 V / 10,000 Ω.
Answer and working
Calculation notes
$I=\frac{U}{R_{\mathrm{total}}}$
$I=\frac{5\,\mathrm{V}}{10000\,\Omega}$
${I=0.0005\,\mathrm{A}=0.5\,\mathrm{mA}}$
What is the output voltage between M and B?
Hint
U_MB = I × R2.
Answer and working
Calculation notes
$U_{MB}=I\cdot R_2$
$U_{MB}=0.0005\,\mathrm{A}\cdot5000\,\Omega$
${U_{MB}=2.5\,\mathrm{V}}$
Now connect R_L = 10 kΩ between M and B. Find R2 parallel R_L.
Hint
R_parallel = R2 × R_L / (R2 + R_L).
Answer and working
Calculation notes
$R_{2\parallel L}=\frac{R_2R_L}{R_2+R_L}$
$R_{2\parallel L}=\frac{5\,\mathrm{k\Omega}\cdot10\,\mathrm{k\Omega}}{5\,\mathrm{k\Omega}+10\,\mathrm{k\Omega}}$
${R_{2\parallel L}\approx3.333\,\mathrm{k\Omega}}$
The dashed load is parallel to the lower resistor.
Calculate the loaded output voltage and check the potentiometer model.
Hint
U_out = 5 V × R_lower / (R1 + R_lower).
Answer and working
Calculation notes
$U_{\mathrm{out}}=U\cdot\frac{R_{2\parallel L}}{R_1+R_{2\parallel L}}$
$U_{\mathrm{out}}\approx5\,\mathrm{V}\cdot\frac{3.333\,\mathrm{k\Omega}}{5\,\mathrm{k\Omega}+3.333\,\mathrm{k\Omega}}$
${U_{\mathrm{out}}=2\,\mathrm{V}}$
The load reduces output voltage. Check the model: 10 kΩ, midpoint, 10 kΩ load.

Think it through before opening the answer. Always write the formula, substituted values and units.

Circuit 05 · 5 small steps

An LED with a series resistor

The LED is forward biased: anode above, cathode below. We use a simplified constant 2 V forward drop. This is not a universal LED value.

Which component limits current?
Hint
The LED itself is not a current limiter.
Answer and working
R1 limits current. Without it, a voltage source can overload the LED.
What voltage appears across R1?
Hint
Source voltage minus LED forward drop.
Answer and working
Calculation notes
$U_R=U-U_F$
${U_R=5\,\mathrm{V}-2\,\mathrm{V}=3\,\mathrm{V}}$
What is the LED current?
Hint
I = U_R / R1.
Answer and working
Calculation notes
$I=\frac{U_R}{R_1}$
$I=\frac{3\,\mathrm{V}}{330\,\Omega}$
${I\approx0.00909\,\mathrm{A}=9.09\,\mathrm{mA}}$
How much power is dissipated in the resistor?
Hint
P_R = U_R × I.
Answer and working
Calculation notes
$P_R=U_R\cdot I$
$P_R\approx3\,\mathrm{V}\cdot0.00909\,\mathrm{A}$
${P_R\approx0.0273\,\mathrm{W}=27.3\,\mathrm{mW}}$
Increase R1 to 680 Ω. Calculate and check the LED model.
Hint
The simplified model still gives 3 V across R1.
Answer and working
Calculation notes
$I=\frac{U-U_F}{R_1}$
$I=\frac{5\,\mathrm{V}-2\,\mathrm{V}}{680\,\Omega}$
${I\approx4.41\,\mathrm{mA}}$
The LED usually becomes dimmer. Brightness is not exactly proportional to current.

Think it through before opening the answer. Always write the formula, substituted values and units.

Circuit 06 · 5 small steps

The diode as a current valve

Anode A_D is above, cathode K below. For hand calculation we assume a 0.7 V forward drop and zero reverse current. The real characteristic is gradual; the interactive diode model uses a different approximation.

Is D1 forward or reverse biased?
Hint
The anode connects to the positive terminal through R1.
Answer and working
Forward biased. The cathode is the side with the bar in the symbol.
Find the voltage across R1.
Hint
Subtract the assumed diode drop from 5 V.
Answer and working
Calculation notes
$U_R=U-U_F$
${U_R=5\,\mathrm{V}-0.7\,\mathrm{V}=4.3\,\mathrm{V}}$
Calculate current.
Hint
I = U_R/R1.
Answer and working
Calculation notes
$I=\frac{U_R}{R_1}$
$I=\frac{4.3\,\mathrm{V}}{1000\,\Omega}$
${I=4.3\,\mathrm{mA}}$
Reverse D1 only. What happens in the model?
Hint
In reverse bias, the ideal diode interrupts the current path.
Answer and working
Calculation notes
$I=0$
$U_R=R_1\cdot I=0\,\mathrm{V}$
${U_{AK}=-5\,\mathrm{V}}$
The negative sign refers to anode relative to cathode. Real diodes have leakage and a maximum reverse voltage.
Why might the plot show a current other than 4.3 mA?
Hint
0.7 V is a calculation assumption, not a fixed law of nature.
Answer and working
The plot uses an exponential characteristic. U_F depends on current, temperature and diode type. Compare the trend and state the model assumptions.

Think it through before opening the answer. Always write the formula, substituted values and units.

Circuit 07 · 5 small steps

Charging a capacitor

C1 is initially uncharged. At t = 0 the ideal source steps from 0 to 5 V. The drawn circuit then stays closed. C1 is ideal; M–B is its voltage.

What is C1 voltage just after switching on?
Hint
Capacitor voltage cannot jump with finite current.
Answer and working
Calculation notes
${u_C(0^+)=0\,\mathrm{V}}$
Capacitor voltage cannot jump with finite current.
What is the initial current?
Hint
Initially the full 5 V appear across R1.
Answer and working
Calculation notes
$i(0^+)=\frac{U-u_C(0^+)}{R}$
$i(0^+)=\frac{5\,\mathrm{V}-0\,\mathrm{V}}{1000\,\Omega}$
${i(0^+)=5\,\mathrm{mA}}$
Calculate the time constant τ.
Hint
1000 µF = 0.001 F; τ = R × C.
Answer and working
Calculation notes
$C=1000\,\mathrm{\mu F}=0.001\,\mathrm{F}$
$\tau=R\cdot C$
$\tau=1000\,\Omega\cdot0.001\,\mathrm{F}$
${\tau=1\,\mathrm{s}}$
Find U_C and I after one time constant.
Hint
U_C = U × (1 − e^−t/τ); I = (U/R) × e^−t/τ.
Answer and working
Calculation notes
$u_C(t)=U\left(1-e^{-\frac{t}{\tau}}\right)$
$u_C(1\,\mathrm{s})=5\,\mathrm{V}\left(1-e^{-\frac{1\,\mathrm{s}}{1\,\mathrm{s}}}\right)$
${u_C(1\,\mathrm{s})\approx3.16\,\mathrm{V}}$
$i(1\,\mathrm{s})=\frac{5\,\mathrm{V}}{1000\,\Omega}\cdot e^{-1}$
${i(1\,\mathrm{s})\approx1.84\,\mathrm{mA}}$
Check: resistor voltage plus capacitor voltage equals the source voltage.
What happens after 5τ, and if R1 doubles?
Hint
After 5τ about 99.3% is reached.
Answer and working
Calculation notes
$u_C(5\tau)=5\,\mathrm{V}\left(1-e^{-5}\right)\approx4.966\,\mathrm{V}$
$i(5\tau)=5\,\mathrm{mA}\cdot e^{-5}\approx0.0337\,\mathrm{mA}$
$\tau_{\mathrm{new}}=2000\,\Omega\cdot0.001\,\mathrm{F}=2\,\mathrm{s}$
${i_{\mathrm{new}}(0^+)=\frac{5\,\mathrm{V}}{2000\,\Omega}=2.5\,\mathrm{mA}}$
The larger resistor slows charging. Final voltage stays the same.

Think it through before opening the answer. Always write the formula, substituted values and units.

Circuit 08 · 5 small steps

An inductor builds up current

At t = 0 the source steps from 0 to 5 V. Initial current is zero. L1 is ideal with no winding resistance. R1 limits the eventual DC current.

What is I just after switching on?
Hint
Inductor current cannot jump with finite voltage.
Answer and working
Calculation notes
${i_L(0^+)=0\,\mathrm{A}}$
$u_R(0^+)=0\,\mathrm{V}$
$u_L(0^+)=5\,\mathrm{V}$
Inductor current cannot jump with finite voltage.
What current is reached eventually?
Hint
An ideal inductor has no voltage drop at steady DC.
Answer and working
Calculation notes
$I_{\infty}=\frac{U}{R}$
$I_{\infty}=\frac{5\,\mathrm{V}}{100\,\Omega}$
${I_{\infty}=0.05\,\mathrm{A}=50\,\mathrm{mA}}$
Calculate τ.
Hint
100 mH = 0.1 H; τ = L/R.
Answer and working
Calculation notes
$L=100\,\mathrm{mH}=0.1\,\mathrm{H}$
$\tau=\frac{L}{R}$
$\tau=\frac{0.1\,\mathrm{H}}{100\,\Omega}$
${\tau=0.001\,\mathrm{s}=1\,\mathrm{ms}}$
What is I at t = τ?
Hint
I = I_∞ × (1 − e^−t/τ).
Answer and working
Calculation notes
$i(t)=I_{\infty}\left(1-e^{-\frac{t}{\tau}}\right)$
$i(1\,\mathrm{ms})=50\,\mathrm{mA}\left(1-e^{-1}\right)$
${i(1\,\mathrm{ms})\approx31.6\,\mathrm{mA}}$
$u_R\approx100\,\Omega\cdot0.0316\,\mathrm{A}=3.16\,\mathrm{V}$
$u_L\approx5\,\mathrm{V}-3.16\,\mathrm{V}=1.84\,\mathrm{V}$
Double L. What changes, and why is sudden disconnection different?
Hint
τ depends on L; I_∞ does not.
Answer and working
Calculation notes
$\tau_{\mathrm{new}}=\frac{0.2\,\mathrm{H}}{100\,\Omega}=2\,\mathrm{ms}$
${I_{\infty}=\frac{5\,\mathrm{V}}{100\,\Omega}=50\,\mathrm{mA}}$
The larger inductor slows current rise. Sudden interruption can create high voltage; the turn-on model does not simulate this turn-off case.

Think it through before opening the answer. Always write the formula, substituted values and units.

Circuit 09 · 5 small steps

RC low-pass: measure across C

An ideal sine source drives R1 and C1. Input A–B, output M–B. The output is unloaded. We compare peak amplitudes in steady state, not switch-on behaviour.

Across which component is output measured?
Hint
Follow M and B.
Answer and working
Across C1. Choosing this output makes the arrangement a low-pass filter.
What do you expect at very low frequency?
Hint
Capacitive reactance rises as frequency falls.
Answer and working
Almost the full input appears across C1: U_out approaches 1 V peak.
Calculate cutoff frequency.
Hint
100 nF = 0.0000001 F; f_c = 1/(2πRC).
Answer and working
Calculation notes
$C=100\,\mathrm{nF}=10^{-7}\,\mathrm{F}$
$f_c=\frac{1}{2\pi RC}$
$f_c=\frac{1}{2\pi\cdot1000\,\Omega\cdot10^{-7}\,\mathrm{F}}$
${f_c\approx1592\,\mathrm{Hz}}$
What are output amplitude and phase at f_c?
Hint
Magnitude is 1/√2; a low-pass has negative phase.
Answer and working
Calculation notes
$|H(f_c)|=\frac{1}{\sqrt{2}}$
$\hat U_{\mathrm{out}}=\frac{1\,\mathrm{V}}{\sqrt{2}}\approx0.707\,\mathrm{V}$
$G=20\log_{10}\left(\frac{1}{\sqrt{2}}\right)\approx-3.01\,\mathrm{dB}$
${\varphi=-45^{\circ}}$
The voltage is not half of the input here.
Double C1. Where does f_c move? Check the filter model.
Hint
f_c is inversely proportional to C.
Answer and working
Calculation notes
$C_{\mathrm{new}}=2C$
$f_{c,\mathrm{new}}=\frac{1}{2\pi R\cdot2C}=\frac{f_c}{2}$
${f_{c,\mathrm{new}}\approx796\,\mathrm{Hz}}$
This calculation assumes no additional output load.

Think it through before opening the answer. Always write the formula, substituted values and units.

Circuit 10 · 5 small steps

RC high-pass: measure across R

C1 is now in the signal path; R1 connects M to B. Input A–B, output M–B. Ideal source, no additional load. Component values and cutoff match the low-pass example.

Across which component is output measured?
Hint
Follow M and B.
Answer and working
Across R1. R and C have swapped positions compared with the low-pass.
What output remains for a constant DC input?
Hint
After charging, no DC flows through the ideal capacitor.
Answer and working
Calculation notes
$I_{\mathrm{DC}}=0$
${U_{\mathrm{out,DC}}=R\cdot I_{\mathrm{DC}}=0\,\mathrm{V}}$
This applies at steady state. A transient output may appear at switch-on.
Calculate f_c.
Hint
f_c = 1/(2πRC), regardless of which component is below.
Answer and working
Calculation notes
$f_c=\frac{1}{2\pi RC}$
$f_c=\frac{1}{2\pi\cdot1000\,\Omega\cdot10^{-7}\,\mathrm{F}}$
${f_c\approx1592\,\mathrm{Hz}}$
The same cutoff does not mean the same response.
Find amplitude and phase at f_c.
Hint
Magnitude is 1/√2; high-pass output leads.
Answer and working
Calculation notes
$\hat U_{\mathrm{out}}=\frac{1\,\mathrm{V}}{\sqrt{2}}\approx0.707\,\mathrm{V}$
$G\approx-3.01\,\mathrm{dB}$
${\varphi=+45^{\circ}}$
Compare high-pass and low-pass at 10 × f_c.
Hint
High-pass: x/√(1+x²); low-pass: 1/√(1+x²), x=f/f_c.
Answer and working
Calculation notes
$x=\frac{f}{f_c}=10$
$|H_{\mathrm{HP}}|=\frac{x}{\sqrt{1+x^2}}=\frac{10}{\sqrt{101}}$
${\hat U_{\mathrm{HP}}\approx0.995\,\mathrm{V}}$
$|H_{\mathrm{LP}}|=\frac{1}{\sqrt{1+x^2}}=\frac{1}{\sqrt{101}}$
${\hat U_{\mathrm{LP}}\approx0.0995\,\mathrm{V}}$
Compare both model curves near 15.9 kHz.

Think it through before opening the answer. Always write the formula, substituted values and units.