Series Circuit Diagram Solver

Read a schematic, turn it into equations, and solve it step by step

A 9 V battery is in series with 470 ohm, 680 ohm and a lamp of 150 ohm. Find every voltage.

Reduce a diagram with 100 ohm in series, then 200 parallel 200, then 50 ohm on 10 V

Check whether a bridge with 1k, 2k, 1.5k and 3k is balanced on 12 V

Identify which resistors in this diagram are in series and which are in parallel

Reading a Schematic as a Set of Nodes

A circuit diagram is a graph, not a picture. Wire length and drawing position mean nothing; only the connections do. Turning a diagram into solvable equations takes three steps.

Label the nodes. Every stretch of unbroken wire is one node at one potential. Give it a name.
Classify each element. Two elements are in series if they share a node with nothing else attached to it. They are in parallel if they share both of their nodes.
Write the relations. Series: same current $I$ in amperes, voltages add. Parallel: same voltage $V$ in volts, currents add.

$R_{\text{series}} = \sum R_n \qquad \frac{1}{R_{\text{parallel}}} = \sum \frac{1}{R_n}$

The operating assumption: an ideal schematic — zero-resistance wire, ideal sources, steady-state DC, and resistive elements only.

The mistake people make: judging series or parallel from how the drawing looks. Two resistors drawn side by side are only parallel if both of their ends genuinely meet at the same two nodes; redraw the diagram if you are unsure.

Solve Inward, Then Expand Back Out

Once classified, collapse the network from the far end toward the source until one resistance remains, get the supply current from $I = V/R_{\text{total}}$ , then walk back out applying $V = IR$ at each stage to recover the individual quantities.

The self-check at the end is Kirchhoff: around any loop the drops must sum to the source voltage, and at any node the currents in must equal the currents out.

$\sum_{\text{loop}} V = 0 \qquad \sum_{\text{node}} I = 0$

Not every diagram reduces. A bridge — four resistors in a diamond with a fifth across the middle — contains no pair that is purely in series or purely in parallel. Test it for balance first:

$\frac{R_1}{R_2} = \frac{R_3}{R_4}$

If the ratios match, the bridge is balanced, the middle branch carries no current, and it can be removed. If they do not, use node voltage analysis or a delta-wye transformation instead of forcing a reduction that does not exist.

Common Mistakes to Avoid

Trusting the drawing over the connections — redraw the schematic with nodes labelled before deciding what is in parallel.
Missing that a wire is a node — components at opposite corners of a diagram joined by plain wire are at the same potential.
Forcing a bridge into series-parallel — check the balance ratio first; if it fails, the reduction does not exist.
Treating a lamp as a fixed resistor — a filament's hot resistance can be ten times its cold value, so its stated operating resistance applies only at its rated voltage.
Ignoring meter loading — a voltmeter across a high-value resistor draws current and changes what you are trying to measure.
Forgetting battery internal resistance — under load the terminal voltage falls below the drawn EMF.
Losing the sign convention in KVL — pick one direction round the loop and stay with it.

示例题目

Step 1: One unbroken loop means one current, so add the resistances:

R = 470 + 680 + 150 = 1300\ \Omega

Step 2:

I = 9.0\ \text{V} \div 1300\ \Omega = 6.923 \times 10^{-3}\ \text{A} = 6.92\ \text{mA}

Step 3:

V_1 = (6.923 \times 10^{-3}\ \text{A})(470\ \Omega) = 3.25\ \text{V}

Step 4:

V_2 = (6.923 \times 10^{-3}\ \text{A})(680\ \Omega) = 4.71\ \text{V}

Step 5:

V_{\text{lamp}} = (6.923 \times 10^{-3}\ \text{A})(150\ \Omega) = 1.04\ \text{V}

Step 6: KVL check:

3.25 + 4.71 + 1.04 = 9.00\ \text{V}

Answer:

I \approx 6.92

mA;

V_1 = 3.25

V_2 = 4.71

V_{\text{lamp}} = 1.04

Step 1:

R_2

and

R_3

share both nodes, so they are parallel:

R_{23} = (200)(200)/(200+200) = 100\ \Omega

Step 2:

R_1

R_{23}

and

R_4

now form a single chain:

R_T = 100 + 100 + 50 = 250\ \Omega

Step 3:

I = 10\ \text{V} \div 250\ \Omega = 0.040\ \text{A} = 40\ \text{mA}

Step 4: Voltage across the parallel section:

V_{23} = (0.040\ \text{A})(100\ \Omega) = 4.0\ \text{V}

Step 5: Each branch:

I_2 = I_3 = 4.0\ \text{V} \div 200\ \Omega = 0.020\ \text{A} = 20\ \text{mA}

Step 6: KCL check:

20 + 20 = 40

mA, matching the supply current

Answer:

R_T = 250

Ω,

I = 40

mA,

V_{23} = 4.0

I_2 = I_3 = 20

Step 1: Balance test:

R_1/R_2 = 1.0/2.0 = 0.500

and

R_3/R_4 = 1.5/3.0 = 0.500

— the ratios match

Step 2: Left midpoint:

V_L = 12\ \text{V} \times 2.0\ \text{k}\Omega \div (1.0 + 2.0)\ \text{k}\Omega = 12 \times 0.6667 = 8.00\ \text{V}

Step 3: Right midpoint:

V_R = 12\ \text{V} \times 3.0\ \text{k}\Omega \div (1.5 + 3.0)\ \text{k}\Omega = 12 \times 0.6667 = 8.00\ \text{V}

Step 4:

V_{\text{bridge}} = V_L - V_R = 8.00 - 8.00 = 0\ \text{V}

Step 5: With zero volts across it, the middle branch carries no current and can be removed, which is what makes the rest reducible

Answer: The bridge is balanced:

V_{\text{bridge}} = 0

V and no current flows in the middle branch

常见问题

Follow the single loop from one battery terminal back to the other. Every component in that loop carries the same current, the resistances add to give the total, and each component's voltage is its own current times its own resistance.

Label every stretch of wire as a node. Two components are in series if they share one node with nothing else attached, and in parallel if they share both of their nodes. Ignore where they sit on the page - only the connections matter.

You are probably looking at a bridge or a multi-source network. Test a bridge for balance with R1/R2 = R3/R4; if it balances, the middle branch carries nothing and can be deleted. Otherwise switch to node voltage or mesh analysis.

Yes. Kirchhoff's voltage law says that going once around a closed loop returns you to the same potential, so the drops must sum to the source EMF. If your numbers do not add up, the arithmetic or the circuit reading is wrong.

Series Circuit Diagram Solver

Read a schematic, turn it into equations, and solve it step by step

Reading a Schematic as a Set of Nodes

Solve Inward, Then Expand Back Out

Common Mistakes to Avoid

示例题目

常见问题

How do I read a series circuit diagram?

How can I tell whether components in a diagram are in series or parallel?

What do I do when a circuit diagram will not reduce?

Do the voltage drops in a series diagram have to add up to the battery voltage?

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