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ScienceQuest
Electricity Practice School

Electricity Practice Problems

Generated circuit practice problems on Ohm’s law, LED resistors, RC time constants and stored energy, marked to within 1.5 percent with the working.

Practice

Question 1 of 40

LED Resistor Calculator

Supply voltage
Vs = 11 V
LED forward voltage
Vf = 3.8 V
Forward current
I = 42 mA
Ω

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Worked answers

The first ten questions from the set above, each with its answer and the working that gets there. The working is carried out in the units each equation takes, so its last line can show the answer before it is converted.

  1. Supply voltage
    Vs = 11 V
    LED forward voltage
    Vf = 3.8 V
    Forward current
    I = 42 mA

    Find the series resistor (R).

    Show the answer and working

    Answer R = 171.4 Ω

    Rearranged R = (Vs − Vf) ÷ I

    1. R = (Vs - Vf) / I
    2. headroom = 11 - 3.8 = 7.2 V
    3. = 7.2 / 0.042 A
    4. = 171.43 ohm

    Check it with the LED Resistor Calculator.

  2. Capacitance
    C = 360 nF
    Time constant
    τ = 1.1 ms

    Find the resistance (R).

    Show the answer and working

    Answer R = 3.056 kΩ

    Rearranged R = τ ÷ C

    1. R = tau / C
    2. = 1.1 ms / 360 nF
    3. = 0.0011 / (3.6 × 10⁻⁷)
    4. = 3055.6 ohm = 3.0556 kohm

    Check it with the RC Time Constant Calculator.

  3. Voltage
    V = 37 V
    Resistance
    R = 400 Ω

    Find the current (I).

    Show the answer and working

    Answer I = 92.5 mA

    Rearranged I = V ÷ R

    1. I = V / R
    2. = 37 / 400
    3. = 0.0925 A = 92.5 mA

    Check it with the Ohm’s Law Calculator.

  4. Capacitance
    C = 64 µF
    Voltage
    V = 10 V

    Find the energy stored (E).

    Show the answer and working

    Answer E = 3.2 mJ

    Rearranged E = ½ C V²

    1. E = 0.5 x C x V^2
    2. = 0.5 x 64 uF x (10 V)^2
    3. = 0.5 x 6.4 × 10⁻⁵ x 100
    4. = 0.0032 J = 3.2 mJ

    Check it with the Capacitor Energy Calculator.

  5. Resistance
    R = 0.4 Ω
    Resistivity
    ρ = 3.8 µΩ·cm
    Cross-sectional area
    A = 2.6 mm²

    Find the length (L).

    Show the answer and working

    Answer L = 27.37 m

    Rearranged L = RA ÷ ρ

    1. L = R x A / rho
    2. = 0.4 ohm x 2.6 mm2 / 3.8 uohm-cm
    3. = 1.04 × 10⁻⁶ / (3.8 × 10⁻⁸)
    4. = 27.368 m

    Check it with the Wire Resistance Calculator.

  6. First charge
    q₁ = 3.8 µC
    Second charge
    q₂ = 0.56 µC
    Separation
    r = 34 cm

    Find the electrostatic force (F).

    Show the answer and working

    Answer F = 0.1654 N

    Rearranged F = k · q₁ · q₂ ÷ r²

    1. F = k q1 q2 / r^2
    2. = 8.988 × 10⁹ x 3.8 × 10⁻⁶ C x 5.6 × 10⁻⁷ C / (0.34 m)^2
    3. = 0.019126 / 0.1156
    4. = 0.16545 N

    Check it with the Coulomb’s Law Calculator.

  7. Input voltage
    Vin = 20 V
    Output voltage
    Vout = 8.5 V
    Top resistor R₁
    3.5 kΩ

    Find the bottom resistor R₂.

    Show the answer and working

    Answer R₂ = 2.587 kΩ

    Rearranged R₂ = R₁ × Vout ÷ (Vin − Vout)

    1. R2 = R1 x Vout / (Vin - Vout)
    2. = 3.5 kohm x 8.5 V / (20 V - 8.5 V)
    3. = 3.5 kohm x 8.5 / 11.5
    4. = 2.587 kohm

    Check it with the Voltage Divider Calculator.

  8. Resistance
    R = 17 kΩ
    Capacitance
    C = 120 nF

    Find the time constant (τ).

    Show the answer and working

    Answer τ = 2.04 ms

    Rearranged τ = R · C

    1. tau = R x C
    2. = 17 kohm x 120 nF
    3. = 17,000 x 1.2 × 10⁻⁷
    4. = 0.00204 s = 2.04 ms

    Check it with the RC Time Constant Calculator.

  9. Voltage
    V = 13 V
    Resistance
    R = 250 Ω

    Find the current (I).

    Show the answer and working

    Answer I = 52 mA

    Rearranged I = V ÷ R

    1. I = V / R
    2. = 13 / 250
    3. = 0.052 A = 52 mA

    Check it with the Ohm’s Law Calculator.

  10. Energy stored
    E = 19 mJ
    Voltage
    V = 39 V

    Find the capacitance (C).

    Show the answer and working

    Answer C = 24.98 µF

    Rearranged C = 2E ÷ V²

    1. C = 2 E / V^2
    2. = 2 x 0.019 / 39^2
    3. = 0.038 / 1521
    4. = 2.498 × 10⁻⁵ F = 24.984 uF

    Check it with the Capacitor Energy Calculator.

Get every prefix into base units first

Circuit values are quoted in the units components are sold in, and the formulas want volts, amps, ohms, farads and seconds. 50 mA is 0.05 A, 10 kΩ is 10,000 Ω, and 100 nF is 1 × 10⁻⁷ F. Do this once at the top of the working.

Then be precise about which voltage a formula wants. Ohm’s law relates the current through a component to the voltage across that component, not the supply voltage. A resistor in series with an LED gets only what is left after the forward drop.

Five ways a circuit answer goes wrong

  • Milliamps treated as amps. 12 V at 50 mA is 12 / 0.05 = 240 Ω. Using 50 A returns 0.24 Ω, out by a factor of a thousand.
  • The whole supply across the series resistor. A 3.2 V white LED at 20 mA from 5 V needs (5 − 3.2) / 0.02 = 90 Ω. Ignoring the forward drop gives 250 Ω and an LED at well under half its intended current.
  • Nanofarads read as microfarads. 10 kΩ with 100 nF is a 1 ms time constant and a 159 Hz cutoff. The same resistor with 100 µF is 1 second and 0.159 Hz.
  • One-way cable length. Current goes out on one core and back on the other, so a load 15 metres away is 30 metres of copper: 0.5034 Ω in 1 mm², dropping 5.03 V at 10 A. The one-way figure halves both.
  • Resistance without dissipation. 12 V at 50 mA is 0.6 W, so the 240 Ω answer is right and a quarter-watt part there will drift and discolour. P = I²R is a second calculation, not a footnote.

Checking by exponent and by anchor

A few anchors catch most slips. 1 kΩ across 5 V passes 5 mA. A capacitor is 63 percent charged after one time constant and 99.3 percent after five, which is why five time constants is the working definition of settled.

Then check the exponents. Power and stored energy both square their voltage or current term, so a 10 percent error in voltage is 21 percent in the energy held by a capacitor. A time constant is linear in both resistance and capacitance. Wire resistance is linear in length and inverse in area, so doubling the diameter quarters it, which matters because cable is sold by diameter.

The part is not its nominal value

The arithmetic here is exact and the components are not. E12 resistors are typically ±10 percent, electrolytic capacitors commonly ±20 percent, and an LED’s forward voltage varies between parts from the same reel. A resistance of 137 Ω is correct and 150 Ω is what you can buy, which is why the calculators round up to the next E12 value: for current limiting, larger is the safe direction.

Resistivity carries its own trap. Materials tables quote it in microohm centimetres at 20 °C, and 1 µΩ·cm is 10⁻⁸ Ω·m, not 10⁻⁶: copper 1.678, aluminium 2.65, nichrome 110. Copper then rises about 0.393 percent per degree Celsius, so a conductor working at 70 °C carries roughly 20 percent more resistance than the table value suggests.

Common questions

Which unit should I type my answer in?

The one shown beside the answer box, which is the unit the matching calculator displays for that quantity. Resistance comes back in ohms or kilohms, current in milliamps, capacitance in microfarads or nanofarads, and a time constant in milliseconds. The marking compares numbers, not units, so an answer of 0.24 where 240 Ω was asked for reads as wrong by a factor of a thousand rather than as the same value in different clothing.

Can I check my working against the calculator?

Yes, and it will always agree. Each question is generated from a calculator specification and solved by that calculator’s own solver, so the two cannot return different answers for the same inputs. Asking for the working shows the substitution the solver performed, which is usually enough to find where a prefix went missing.