A battery drawn on a circuit is not guaranteed to supply energy. Another source can push current back through it. Signed equations let the circuit tell us what happens.
We move from node labels and loop signs to loaded dividers, opposing sources and unbalanced bridges.
What you will learn to do
- Identify nodes and valid series/parallel reductions.
- Use negative current results consistently.
- Solve multibranch networks through node voltages or loop currents.
- Distinguish an ideal source from a source with series resistance.
- Verify solutions with current, voltage and power balances.
Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Kirchhoff networks and current direction Learning goals: - Identify nodes and valid series/parallel reductions. - Use negative current results consistently. - Solve multibranch networks through node voltages or loop currents. - Distinguish an ideal source from a source with series resistance. - Verify solutions with current, voltage and power balances. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
BEFORE YOU READ
Bring back one idea
Try a sentence or a sketch before opening the comparison. If this is new, use the explanation to get started.
Why can drawing two resistors next to each other fail to put them in series?
Compare your reasoning
The electrical connections decide current paths. If their shared node has another branch, current can split; the two resistor currents need not match.
Revisit the explanation →01 / Kirchhoff networks and current direction
Read connections before reading the drawing
A circuit drawing is a map of connections, not a photograph. Stretch an ideal wire, fold it around a corner or redraw it straight: points joined by that wire remain one node. Resistors that look adjacent are not automatically in series. They share a current only when their intervening node has no other current-carrying branch.
Two components are in parallel when both of their endpoints connect to the same pair of nodes. Equal-looking shapes or equal resistance values do not establish that connection. Label the nodes before trying to simplify a crowded drawing; the labels survive every redraw.
For this chapter, wires have negligible resistance, sources and resistors are lumped components, currents have settled, and no changing external magnetic flux links the loops. Real wire resistance, capacitance and induction can matter in other regimes. The simple model is useful because we state what it leaves out.
| Observation | Consequence | Check first |
|---|---|---|
| One shared intermediate node, no extra branch | Same current through the two components | Could current leave through a meter or load? |
| Both endpoints share the same two nodes | Same potential difference | Are all apparent wires actually connected? |
| A resistor joins the middle of two branches | Usually no direct series/parallel reduction | Can symmetry or node equations determine its current? |
WORKED EXAMPLE
A voltmeter becomes part of the divider
An ideal 12 V source feeds a 3 kΩ upper resistor and a 6 kΩ lower resistor in series. A 6 kΩ voltmeter is connected across the lower resistor. What does it read?
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Kirchhoff networks and current direction Focus: Read connections before reading the drawing Learning goals: - Identify nodes and valid series/parallel reductions. - Use negative current results consistently. - Solve multibranch networks through node voltages or loop currents. - Distinguish an ideal source from a source with series resistance. - Verify solutions with current, voltage and power balances. Equations in this section (LaTeX): Build the physical reasoning before choosing an equation. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
02 / Kirchhoff networks and current direction
Current arrows are a bookkeeping choice
If charge entered a node faster than it left indefinitely, charge would accumulate there and the voltages would change. A settled direct-current circuit has no such continuing accumulation. Its incoming current sum therefore equals its outgoing current sum.
Choose one reference direction for each branch. You do not have to know the actual direction in advance, and the arrows do not need to imitate traffic that looks plausible. A negative answer means the actual conventional current runs opposite to its arrow. Keep that result in the equations; changing a drawing halfway through without changing the signs is a common way to lose consistency.
The more general account is incoming minus outgoing current equals the rate of charge accumulation. At a charging capacitor plate that rate is not zero. At an ideal wire node in our settled circuit it is zero. Charge conservation is the principle; the zero-accumulation equation is its stated special case.
The last equality applies to the settled node model. The ampere is coulomb per second.
WORKED EXAMPLE
Let the sign correct your first guess
At a settled node, 0.80 A enters and 1.10 A leaves through two known branches. A third current I is initially drawn leaving. Find I.
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STOP & THINK
A solved branch current is negative. What should you do first?
- Replace it by zero
- Interpret flow opposite to its chosen arrow
- Declare the circuit impossible
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Kirchhoff networks and current direction
Focus: Current arrows are a bookkeeping choice
Learning goals:
- Identify nodes and valid series/parallel reductions.
- Use negative current results consistently.
- Solve multibranch networks through node voltages or loop currents.
- Distinguish an ideal source from a source with series resistance.
- Verify solutions with current, voltage and power balances.
Equations in this section (LaTeX):
\sum I_{\rm in}-\sum I_{\rm out}=\frac{dQ_{\rm node}}{dt}=0
Meaning: The last equality applies to the settled node model. The ampere is coulomb per second.
Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.
Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.03 / Kirchhoff networks and current direction
Walk around a loop and keep a voltage ledger
Start at a node and walk around a closed circuit path. Returning to that node must return to its potential. Record the potential rise or fall across each component with the direction of your walk, not with an assumed universal plus sign for batteries and minus sign for resistors.
For a positive resistance, conventional current flows from higher to lower potential. Traversing in the chosen positive current direction contributes −IR; traversing against it contributes +IR. These rules still work when the solved I is negative. Across an ideal source, going from its labelled negative terminal to its positive terminal contributes +emf, regardless of whether the source is delivering or absorbing power.
The branch current direction and the loop-walk direction are separate choices. Keeping them separate lets you reverse a loop equation without changing its physical meaning: every term changes sign together. With changing flux, induced emf must be included explicitly; the simple electrostatic walk alone would be incomplete.
| Element traversed | Contribution |
|---|---|
| Resistor along the reference current | −IR |
| Resistor against the reference current | +IR |
| Ideal source from − terminal to + terminal | +emf |
| Ideal source from + terminal to − terminal | −emf |
WORKED EXAMPLE
One source pushes against the other
A single loop contains 18 V and 6 V ideal sources opposing each other, plus 2 Ω and 4 Ω resistors. Choose current in the direction driven by the 18 V source. Find the current.
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Kirchhoff networks and current direction Focus: Walk around a loop and keep a voltage ledger Learning goals: - Identify nodes and valid series/parallel reductions. - Use negative current results consistently. - Solve multibranch networks through node voltages or loop currents. - Distinguish an ideal source from a source with series resistance. - Verify solutions with current, voltage and power balances. Equations in this section (LaTeX): Build the physical reasoning before choosing an equation. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
04 / Kirchhoff networks and current direction
One unknown voltage can replace several current guesses
A node-voltage method chooses one node as zero and expresses every branch current through the remaining node potentials. Ohm’s law then turns a current balance into an equation for those potentials. The method is especially economical when several branches connect between the same two rails.
Consider two ideal sources, each in series with its own resistance, connected between a common bottom rail and a common top rail. Both source positive terminals point toward the top rail through their resistors. Call the top voltage V and the bottom voltage zero. The currents delivered upward by the source branches are (E₁−V)/r₁ and (E₂−V)/r₂. A load R between the rails takes V/R downward.
The source emfs need not equal the rail voltage because current causes a drop across each series resistance. If V exceeds a source’s emf, its signed delivery current becomes negative. It is receiving current at its positive terminal. The same equation handles this reversal without a second set of rules.
Both source currents are positive when delivered toward the top rail. The load current is positive from top to bottom.
Explain this equation
Read it in words
The currents supplied by the two source branches add to the current leaving through the load resistor.
Symbols and units
- Signed source emfs in the illustrated orientation; these E symbols are emfs, not electric fields.Volts (V).
- Potential of the top node relative to the common bottom node.Volts (V).
- Positive internal resistances of the two source branches.Ohms (Ω).
- Positive load resistance connected across the common nodes.Ohms (Ω).
- Current in source branch j, taken positive when it supplies the top node.Amperes (A).
Signs and operations
Each numerator is the voltage available across that branch’s resistance in the assumed current direction. A negative result means the current actually flows the other way; it is not an invalid answer.
When it applies
Use the shown parallel-source topology in steady state, with the stated source polarities, linear resistors and negligible charge accumulation at a node. Do not apply this particular equation to a different wiring diagram.
Make it concrete
For E₁ = 12 V, E₂ = 6 V, r₁ = r₂ = 2 Ω and R = 4 Ω, V = 7.2 V. Currents are +2.4 A, −0.6 A and 1.8 A: the weaker source absorbs current while the stronger one supplies both branches.
Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Equation (LaTeX): \frac{E_1-V}{r_1}+\frac{E_2-V}{r_2}=\frac{V}{R}
Read it as: The currents supplied by the two source branches add to the current leaving through the load resistor.
Symbols:
E_1,\ E_2: Signed source emfs in the illustrated orientation; these E symbols are emfs, not electric fields.; Volts (V).
V: Potential of the top node relative to the common bottom node.; Volts (V).
r_1,\ r_2: Positive internal resistances of the two source branches.; Ohms (Ω).
R: Positive load resistance connected across the common nodes.; Ohms (Ω).
\frac{E_j-V}{r_j}: Current in source branch j, taken positive when it supplies the top node.; Amperes (A).
Signs and operations: Each numerator is the voltage available across that branch’s resistance in the assumed current direction. A negative result means the current actually flows the other way; it is not an invalid answer.
Conditions: Use the shown parallel-source topology in steady state, with the stated source polarities, linear resistors and negligible charge accumulation at a node. Do not apply this particular equation to a different wiring diagram.
Help me explain the equation in words, predict how changing a quantity affects the result, and apply it to a small numerical example.
Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.WORKED EXAMPLE
A weaker source can absorb energy
Use E₁ = 12 V, r₁ = 2 Ω, E₂ = 6 V, r₂ = 3 Ω and load R = 6 Ω in the described two-rail network. Find V and all branch currents.
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Kirchhoff networks and current direction
Focus: One unknown voltage can replace several current guesses
Learning goals:
- Identify nodes and valid series/parallel reductions.
- Use negative current results consistently.
- Solve multibranch networks through node voltages or loop currents.
- Distinguish an ideal source from a source with series resistance.
- Verify solutions with current, voltage and power balances.
Equations in this section (LaTeX):
\frac{E_1-V}{r_1}+\frac{E_2-V}{r_2}=\frac{V}{R}
Meaning: Both source currents are positive when delivered toward the top rail. The load current is positive from top to bottom.
Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.
Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.05 / Kirchhoff networks and current direction
Change a source and predict when its current reverses
Before touching a control, predict what would make the second branch current zero. Its source must match the rail voltage, but the rail voltage itself is set by the rest of the circuit. At that point the second branch can be removed from the current balance and the first source with the load forms a simple divider.
The explorer keeps every resistance positive and solves the same settled model after each change. The arrows show the actual directions; the signed readouts keep the original upward source references. Nothing represents a transient charging curve. Compare the currents and power account after each manual change.
At R = 6 Ω, the no-current condition for the second source is E₂ = 12×6/(2+6) = 9 V. Below that value it absorbs current; above it it supplies current. Changing R moves the boundary. A memorized statement that “the smaller battery charges” is therefore less useful than comparing each emf with the solved terminal voltage.
INTERACTIVE LAB
Two sources, one shared voltage
Start with R = 6 Ω. Predict the second source’s current at 6 V, 9 V and 12 V, then use the control to check. Compare total signed source power with all resistor heating.
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STOP & THINK
With the first source 12 V, its series resistance 2 Ω and load 4 Ω, what second-source emf gives zero current in its branch?
Answer in V.
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Kirchhoff networks and current direction Focus: Change a source and predict when its current reverses Learning goals: - Identify nodes and valid series/parallel reductions. - Use negative current results consistently. - Solve multibranch networks through node voltages or loop currents. - Distinguish an ideal source from a source with series resistance. - Verify solutions with current, voltage and power balances. Equations in this section (LaTeX): Build the physical reasoning before choosing an equation. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
06 / Kirchhoff networks and current direction
When symmetry is absent, keep both middle nodes
An unbalanced bridge has a resistor joining the middle nodes of two dividers. The link changes both divider voltages, so calculating the two unloaded divider voltages and treating their difference as the final answer generally fails. Use them only to predict a likely direction before solving the loaded network.
Choose the bottom rail as zero and the top rail as the supplied voltage. Label the left middle potential VA and the right middle potential VB. Write a current balance at each. The bridge current then follows from (VA−VB)/Rbridge, with its direction emerging from the sign.
A balanced bridge is a special simplification, not the default. If the unlinked divider voltages are equal, connecting a passive bridge resistor does not change them and its current is zero. Unequal divider voltages require the coupled equations.
WORKED EXAMPLE
Solve the link instead of deleting it
In the diagram, the left upper/lower resistors are 3 Ω and 6 Ω; the right upper/lower resistors are 6 Ω and 3 Ω. A 6 Ω resistor joins the middle nodes. Find its current with a 15 V ideal supply.
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Kirchhoff networks and current direction Focus: When symmetry is absent, keep both middle nodes Learning goals: - Identify nodes and valid series/parallel reductions. - Use negative current results consistently. - Solve multibranch networks through node voltages or loop currents. - Distinguish an ideal source from a source with series resistance. - Verify solutions with current, voltage and power balances. Equations in this section (LaTeX): Build the physical reasoning before choosing an equation. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
07 / Kirchhoff networks and current direction
A solved circuit must also pay its energy bill
A set of equations can be solved accurately even if one of those equations has the wrong sign. A separate power account is an efficient check. A positive resistor always dissipates I²R, whatever reference direction you chose. An ideal source’s signed delivered power is EI when I is referenced leaving its positive terminal.
For the two-source network, add both signed ideal-source powers. One may be negative because that source stores or otherwise absorbs energy in its ideal model. The result must equal heating in both series resistors and the load. Do not compare only the stronger source with the load and forget the weaker source or internal heating.
Power is energy per time, not an extra conservation law independent of energy. Multiplying a correctly signed loop equation by a suitable current can recover the same balance. Checking it by summing every physical component is still valuable because it catches omitted branches and sign errors.
This identity applies to the declared two-source resistive model. Negative source power denotes net absorption by that ideal emf element.
WORKED EXAMPLE
Account for the source that receives current
For the earlier 12 V/2 Ω, 6 V/3 Ω and 6 Ω load network, verify the total power balance.
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Kirchhoff networks and current direction
Focus: A solved circuit must also pay its energy bill
Learning goals:
- Identify nodes and valid series/parallel reductions.
- Use negative current results consistently.
- Solve multibranch networks through node voltages or loop currents.
- Distinguish an ideal source from a source with series resistance.
- Verify solutions with current, voltage and power balances.
Equations in this section (LaTeX):
E_1I_1+E_2I_2=I_1^2r_1+I_2^2r_2+\frac{V^2}{R}
Meaning: This identity applies to the declared two-source resistive model. Negative source power denotes net absorption by that ideal emf element.
Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.
Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.08 / Kirchhoff networks and current direction
Use enough independent equations, then test the model
Writing every possible loop is rarely necessary. In a connected network, one node-current equation is redundant because summing all of them cancels every internal branch. Additional loop equations can also repeat information. The goal is enough independent equations for the unknowns, not the largest pile of algebra.
Pick node voltages when many branches share a few junctions. Pick loop currents when the drawing has a small number of simple meshes; a shared resistor then carries a difference of mesh currents. Keep an explicit polarity reference for any measured voltage. Neither method changes the circuit.
After solving, substitute into an unused node or loop, check source and resistor powers, and consider a limit. If a bridge resistance becomes very large, its current should tend to zero. If you short two nodes with an ideal wire, their voltage difference must vanish; an ideal current can still flow through that wire.
WORKED EXAMPLE
A shared resistor carries a difference of mesh currents
Two clockwise mesh currents satisfy 5i₁ − 2i₂ = 11 and −2i₁ + 6i₂ = 6, in volts with resistances in ohms. The shared 2 Ω branch is traversed oppositely by the meshes. Find its current in mesh one’s direction.
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Kirchhoff networks and current direction Focus: Use enough independent equations, then test the model Learning goals: - Identify nodes and valid series/parallel reductions. - Use negative current results consistently. - Solve multibranch networks through node voltages or loop currents. - Distinguish an ideal source from a source with series resistance. - Verify solutions with current, voltage and power balances. Equations in this section (LaTeX): Build the physical reasoning before choosing an equation. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
BRING THE IDEAS TOGETHER
What should stay with you
- Label connections before simplifying a drawing.
- Current arrows are reference choices; a negative answer reverses the physical flow.
- Keep source polarity separate from loop traversal and branch-current direction.
- Check both charge conservation and the complete energy account.
RETRIEVE, THEN EXPLAIN
Try it without the examples
Work these out before checking. If an answer surprises you, identify which assumption or step needs another look. A correct choice is most useful when you can explain why the alternatives fail.
SELF CHECK 1
Two points are connected by an ideal wire. Which must be true in this settled model?
- Their current is zero
- Their potential difference is zero
- Their total charges are equal
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SELF CHECK 2
Mesh currents of 4.5 A and 1.5 A traverse a shared resistor oppositely. What is the shared current in the first mesh direction?
Answer in A.
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SELF CHECK 3
An ideal source has negative signed delivered power. What does this mean?
- It is absorbing energy
- Its series resistor has negative heating
- Its emf must be zero
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APPLY THE CHAPTER
Take the ideas into a problem
Try a complete problem next. Commit to a method before revealing a hint, then compare your reasoning with the worked solution.
The resistor receives the remaining voltageIntermediate · Free sample · Hints and guided reasoningAn apparently balanced divider is loadedIntermediate · Hints and guided reasoningWhich source receives the energy?Intermediate · Hints and guided reasoningFind what the supply sees through a bridgeAdvanced · Hints and guided reasoningFind the emf that makes a branch quietAdvanced · Hints and guided reasoningInfer a source from a reversed branch currentIntermediate · Hints and guided reasoningCheck mesh algebra with source powerAdvanced · Hints and guided reasoningChoose a physically available cell arrangementAdvanced · Hints and guided reasoningChoose efficiency instead of peak powerIntermediate · Hints and guided reasoningOpen terminals can hide an internal currentAdvanced · Hints and guided reasoningPredict a load through an equivalent sourceAdvanced · Hints and guided reasoningSimplify the branches before finding the current.Intermediate · Hints and guided reasoningInsert a dielectric after disconnectingIntermediate · Hints and guided reasoningThe load sees less than the emfIntermediate · Hints and guided reasoningDoubling voltage changes heatingIntermediate · Hints and guided reasoningChoose a current-limiting resistorIntermediate · Hints and guided reasoningWhere does the spare current go?Intermediate · Hints and guided reasoningLeave enough current for regulationIntermediate · Hints and guided reasoningCOME BACK TO THE IDEA
What can you explain later?
Try these from memory at your next study session, then again a few days later. Say your reasoning aloud or sketch it before looking. Adjust the gap to what you can explain.
Next study session
A source has negative delivered current in the solved network. What happened to its role?
Compare your reasoning
Current enters its positive terminal, so its ideal emf absorbs energy under the declared reference. The result does not make the associated positive resistor cool; it still dissipates I²R.
Revisit the explanation →A few days later: change the situation
Can you always find an unbalanced bridge current by subtracting the two unloaded divider voltages and dividing by the bridge resistance?
Compare your reasoning
No. The cross-link loads both dividers and changes their middle-node voltages. Solve the coupled node equations or include the rest of the network’s equivalent resistance.
Revisit the explanation →Compare the physical reason, the assumptions and the conclusion. Your wording can differ. If one part is missing, revisit that section and try explaining it again after a break.
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