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LNGR SPKRHIFI. TWO WAYS. HANDCRAFTED.
LNGR · Learning laboratory

See current flow and voltage drop

Step by step: understand, predict, adjust and calculate.

← Learning laboratory

01 · Flow and difference in level

Moving dots represent conventional current, not electrons at a realistic speed. Voltage bars instead show potential relative to the return. A tall bar does not automatically imply high current: resistance along the path matters too.

02 · A resistor does not use up charge

At steady state, the same current flows before and after a resistor with no branch. Charge does not disappear. The resistor converts energy, so electric potential decreases in the direction of conventional current. Voltage is always measured between two points.

03 · Change one variable, watch two

Increase R₁ while holding the source fixed. Total current falls, and less voltage remains across the parallel branches. Watch current and voltage together. Repeat with R₂: this also changes how current splits between branches.

04 · The meter becomes part of the circuit

An ideal voltmeter does not load the circuit. A real meter has input resistance, which can alter a high-resistance divider. An ammeter must be inserted into the current path; connecting it across the source creates a dangerously low-resistance branch.

Workshop takeaway

Do not just ask “How many volts?” Ask “Between which points?” For current, ask “Through which branch?”

20 exercises · PDF ↓Teacher solutions · PDF ↓

Work it out · answers unfold

Write down your prediction first, then your calculation. Open the answer only when you are ready to compare.

01 · An ideal 6 V source supplies 300 Ω. Mark conventional current direction and find I.

Show worked answer
  1. Outside the source: positive to negative.
  2. Divide voltage by resistance.
  3. I = 20 mA.
Calculation
I=\frac{6\,V}{300\,\Omega}=0.020\,A

02 · Convert 0.004 A to mA and 4.7 kΩ to Ω. Why use consistent units?

Show worked answer
  1. Milli means one thousandth: 4 mA.
  2. Kilo means one thousand: 4700 Ω.
  3. Convert units before substitution.
Calculation
0.004\,A=4\,mA;\quad 4.7\,k\Omega=4700\,\Omega

03 · A 470 Ω resistor has 9 V across it. Find current and dissipation. Is 0.125 W sufficient?

Show worked answer
  1. I = 9/470 = 19.15 mA.
  2. P = 81/470 = 0.172 W.
  3. 0.125 W is insufficient; choose a rating with margin.
Calculation
P=\frac{U^2}{R}=\frac{81}{470}\,W\approx0.172\,W

04 · 220 Ω and 330 Ω are in series across 11 V. Find total resistance and current.

Show worked answer
  1. Add series resistors: 550 Ω.
  2. Both carry the same current.
  3. 11 V / 550 Ω = 20 mA.
Calculation
R_\Sigma=220+330=550\,\Omega

05 · Use the previous series circuit. Find both voltage drops and check their sum.

Show worked answer
  1. I = 20 mA.
  2. U1 = 0.02 · 220 = 4.4 V; U2 = 6.6 V.
  3. 4.4 V + 6.6 V = 11 V.
Calculation
U_1+U_2=4.4\,V+6.6\,V=11\,V

06 · Two 1 kΩ resistors are in parallel. Find equivalent resistance.

Show worked answer
  1. Both branches share voltage.
  2. Equal parallel resistors halve the resistance.
  3. Result: 500 Ω.
Calculation
R_p=\frac{1000\cdot1000}{1000+1000}=500\,\Omega

07 · 300 Ω and 600 Ω are parallel across 6 V. Find branch and total currents.

Show worked answer
  1. I1 = 20 mA; I2 = 10 mA.
  2. Currents add at the node.
  3. Total current: 30 mA.
Calculation
I_\Sigma=\frac{6}{300}+\frac{6}{600}=0.030\,A

08 · Without calculating: can 300 Ω in parallel with 600 Ω exceed 300 Ω? Explain.

Show worked answer
  1. No.
  2. The extra branch allows more current at the same voltage.
  3. Equivalent resistance is below the smallest branch resistance.

09 · Mixed: R1 = 100 Ω in series with R2 = 300 Ω parallel R3 = 600 Ω; source 12 V. Find total R.

Show worked answer
  1. First replace the parallel group: 200 Ω.
  2. Then add R1.
  3. Total R = 300 Ω.
Calculation
R_\Sigma=100+\frac{300\cdot600}{300+600}=300\,\Omega

10 · Use the mixed circuit in task 9. Find total current and parallel-group voltage.

Show worked answer
  1. Total I = 12/300 = 40 mA.
  2. U1 = 0.04 · 100 = 4 V.
  3. R2 and R3 each have 8 V across them.
Calculation
U_p=12\,V-4\,V=8\,V

11 · Find both branch currents in task 9. Verify Kirchhoff’s current law.

Show worked answer
  1. I2 = 8/300 = 26.67 mA.
  2. I3 = 8/600 = 13.33 mA.
  3. Sum = 40 mA = source current.
Calculation
I_2+I_3=26.67+13.33=40\,mA

12 · R3 in task 9 becomes open-circuit. Find the new source current.

Show worked answer
  1. Only R1 and R2 remain as a current path.
  2. Total R = 100 + 300 = 400 Ω.
  3. Total I = 30 mA; current decreases.
Calculation
I=\frac{12}{400}=0.030\,A

13 · R3 in task 9 is ideally shorted. What are R2 voltage and source current?

Show worked answer
  1. The parallel group is bypassed.
  2. U2 = 0 V and I2 = 0 A.
  3. Only R1 limits current: total I = 120 mA.
Calculation
I_\Sigma=\frac{12}{100}=0.120\,A

14 · Divider: R1 = 2 kΩ upper, R2 = 3 kΩ lower, input 10 V. Unloaded output across R2. Find output.

Show worked answer
  1. Total resistance: 5 kΩ.
  2. Current: 2 mA.
  3. Voltage across R2: 6 V.
Calculation
U_o=10\frac{3}{2+3}=6\,V

15 · Load the divider in task 14 with 3 kΩ across R2. Find the new output.

Show worked answer
  1. R2 parallel load = 1.5 kΩ.
  2. Total resistance: 3.5 kΩ.
  3. Output = 10 · 1.5/3.5 = 4.286 V.
Calculation
U_o=10\frac{1.5}{2+1.5}\approx4.286\,V

16 · An unloaded divider should turn 12 V into 3 V. Choose R2 = 1 kΩ. Find R1.

Show worked answer
  1. Output is one quarter of input.
  2. Total resistance must be four times R2.
  3. R1 = 4 kΩ - 1 kΩ = 3 kΩ.
Calculation
R_1=R_2\left(\frac{U_i}{U_o}-1\right)=3\,k\Omega

17 · An 8 Ω resistor has 4 V DC across it. Find P. What happens at 8 V?

Show worked answer
  1. At 4 V: P = 16/8 = 2 W.
  2. At 8 V: P = 64/8 = 8 W.
  3. Double voltage means four times the power.
Calculation
P=\frac{U^2}{R}

18 · A 1 kΩ resistor has ±5% tolerance. Find minimum and maximum current at ideal 5 V.

Show worked answer
  1. Rmin = 950 Ω, Rmax = 1050 Ω.
  2. Minimum I uses maximum R: 4.762 mA.
  3. Maximum I uses minimum R: 5.263 mA.
Calculation
I_{min}=\frac{5}{1050};\quad I_{max}=\frac{5}{950}

19 · How do you connect ideal ammeter and voltmeter to a resistor? Explain the short-circuit mistake.

Show worked answer
  1. Ammeter in series with the current path.
  2. Voltmeter in parallel across the resistor.
  3. An ammeter across the source nearly shorts it.

20 · Verify the power balance of task 9: source power and three resistor dissipations.

Show worked answer
  1. Source: 12 · 0.04 = 0.48 W.
  2. R1: 0.16 W; R2: 64/300 = 0.2133 W; R3: 64/600 = 0.1067 W.
  3. Sum = 0.48 W.
Calculation
P_Q=P_1+P_2+P_3=0.48\,W

Further reading

Original teaching models and authored exercises. Sources provide technical background; no manufacturer measurement curves are reproduced.