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

Untangle mixed circuits

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

← Learning laboratory

01 · Find the nodes first

A circuit can look like a road map. What matters is not whether a resistor is drawn left or right, but which nodes it connects to. Every point on an ideal uninterrupted wire has the same potential. Mark the nodes first. Two components are parallel when both terminals connect to the same two nodes.

02 · Simplify from the inside out

R₁ is before a branch in this example. R₂ and R₃ then share two nodes. Combine those two into an equivalent resistance first. Only then add R₁. The equivalent is a calculation tool: it replaces the two physical resistors only as seen from outside.

03 · Work back from total current

Once total resistance is known, calculate source current. All of it passes through R₁. Its voltage drop tells you how much voltage remains across the parallel section. Now work inward again: divide that same node voltage by each branch resistance to get the two branch currents.

04 · Use three independent checks

Branch currents must add up to total current. Voltage drops along a complete path must add up to source voltage. Finally, total resistor dissipation must equal power supplied by the source. If a check fails, first look for confusion between total voltage and branch voltage.

05 · Not every network reduces this way

A bridge network may contain resistors that are neither simply series nor parallel. Node voltages or loop equations are then useful. “They are drawn one after another” is not a rule for series connection. This route starts with reducible networks, then the exercises progress to a loaded voltage divider.

Workshop takeaway

Mark nodes, simplify, then work back through the original circuit. Adding the branch currents at the end catches many mistakes without an extra instrument.

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.