Difficulty: Intermediate Prerequisites: Ohm's Law, series and parallel resistors, basic KVL and KCL, matrix/simultaneous equation solving.
Most real circuits are not single loops. The moment you have more than one path for current, you need a systematic approach: Kirchhoff's laws. This unit pushes you from "one loop, one equation" to "multiple loops, multiple unknowns," which is where the algebra starts to matter as much as the physics. The Wheatstone bridge is a classic special case that turns up in sensor circuits and measurement instruments. The problems here also introduce internal resistance of batteries, current dividers in parallel networks, and power dissipation, all of which recur throughout the rest of the course.
For circuits with multiple loops, assign current variables, write KCL equations at junctions and KVL equations around loops, then solve the system of equations simultaneously. The Wheatstone bridge is a four-resistor diamond circuit where, at balance, no current flows through the bridge element, and you can find an unknown resistance from the other three.
Kirchhoff's Current Law (KCL)
At any junction (node) in a circuit, the sum of currents entering equals the sum of currents leaving. Equivalently, the algebraic sum of all currents at a node is zero.
In simple terms, charge does not pile up at a junction. Everything that flows in must flow out.
Kirchhoff's Voltage Law (KVL)
Around any closed loop in a circuit, the sum of all voltage gains and drops is zero. You pick a direction, walk around the loop, and add up every EMF and every IR drop with the correct sign.
In simple terms, if you walk in a circle through a circuit and add up all the voltage changes, you end up back where you started, so the total is zero.
Wheatstone bridge
A diamond-shaped circuit with four resistors and a galvanometer (or a branch element) across the middle. When the bridge is "balanced," no current flows through the middle branch, and the unknown resistance can be calculated from the ratio of the other three.
Internal resistance (r)
The resistance inside a real battery or voltage source. It causes the terminal voltage to be lower than the EMF when current flows: V_terminal = EMF − I × r.
In simple terms, a real battery wastes some of its voltage pushing current through its own insides.
Current divider
When current reaches a parallel split, it divides inversely with resistance. For two parallel resistors: I₁ = I_total × R₂ / (R₁ + R₂). The smaller resistor carries the larger share.
Power dissipation
The rate at which a resistor converts electrical energy to heat: P = I²R = V²/R = IV. Measured in watts.
Step 1 – Label currents. Assign a current variable and an assumed direction to every branch. If your assumed direction is wrong, the answer will simply come out negative.
Step 2 – Write KCL at each independent junction. If the circuit has N junctions, you get N − 1 independent KCL equations (the last one is redundant).
Step 3 – Write KVL around each independent loop. Walk around the loop in your chosen direction. For a resistor: if you walk in the direction of current, write −IR (voltage drop); against current, write +IR. For a battery: if you cross from − to +, write +EMF; from + to −, write −EMF.
Step 4 – Solve the system. You need as many independent equations as you have unknown currents. Substitute and solve, or use matrices.
When traversing a loop for KVL:
Crossing a resistor in the direction of the assigned current: voltage drop, so −IR.
Crossing a resistor against the assigned current: voltage rise, so +IR.
Crossing a battery from − to +: voltage rise, so +EMF.
Crossing a battery from + to −: voltage drop, so −EMF.
Consistency matters more than which convention you pick. Stick with one and do not switch mid-problem.
Circuit: six resistors (R₁ = R₅ = 55 Ω, R₂ = R₆ = 148 Ω, R₃ = 84 Ω, R₄ = 76 Ω) with two voltage sources (V₁ = 18 V, V₂ = 12 V). Five branch currents: I₁ through I₅.
KCL at the central node gives: I₂ = I₁ + I₃.
KVL around the left loop: 0 = V₁ − I₃R₃ + I₁R₁ (signs depend on direction choices).
KVL around the right loop: 0 = I₃R₃ − I₄R₄ − V₂.
KVL around the outer loop: 0 = I₂R₂ − V₁.
After substitution, the homework finds:
I₃ = (I₁R₁ − V₁) / R₃
I₂ = (V₂ − I₁R₁) / (2R₂)
Combining: 0.0405 − 0.185 I₁ = 1.654 I₁ − 0.214, giving I₁ = 0.138 A.
Then I₂ = (12 − 0.138 × 55) / (2 × 148) = 0.0148 A.
The key takeaway is the method: label, write KCL, write KVL, substitute, solve.
When you need the voltage at a particular node in a parallel network, use the voltage divider or the current divider to get there.
For the circuit on the last page of the homework: R₁ = R₃ = 28 Ω, R₄ = R₅ = 90 Ω, R₂ = 82 Ω, V_bat = 11.58 V, V_emf = 12 V.
R₃₄₅ (the series combination of R₃ in series with the parallel combination of R₄ and R₅) collapses to a single equivalent.
R₂ is in parallel with R₃₄₅ to give R₂₃₄₅ = (R₂ × R₃₄₅) / (R₂ + R₃₄₅) = 59.813 Ω.
R_eq = R₁ + R₂₃₄₅ = 86.81 Ω.
Total current: I₁ = V_bat / R_eq = 11.58 / 86.81 = 0.133 A.
Internal resistance: r = (V_emf − V_bat) / I₁ = (12 − 11.58) / 0.133 = 3.15 Ω.
Current divider splits I₁ between R₂ and R₃₄₅: I₂ = I₁ × R₃₄₅ / (R₂ + R₃₄₅) = 0.075 A.
Voltage across R₂: V₂ = I₂ × R₂ = 0.075 × 82 = 7.79 V... though the exact numbers depend on the precise R values.
Power in R₂: P₂ = I₂² × R₂.
The classic Wheatstone bridge has four resistors in a diamond arrangement with a voltage source across one diagonal and a detector (galvanometer) across the other.
Given: R₁ = 41 Ω, R₂ = 122 Ω, R₃ = 116 Ω, R₄ = 104 Ω, R_x = ?, V = 12 V. The bridge is balanced (I₄ = 0 through the galvanometer branch).
When balanced, no current flows through the bridge branch (I₄ = 0).
This means I₁ = I₃ (current through the top path) and I₂ = I₅ (current through the bottom path). Wait, let me re-read the circuit. In this problem, R₁ and R₃ form one path, and R₂ and R₄ form the other, with R_x across the bridge.
At balance: R₁/R₃ = R₂/R_x (the bridge balance condition), or equivalently R_x = R₂ × R₃ / R₁.
I₁ = V / (R₁ + R₃) = 12 / (41 + 116) = 0.0764 A.
V₁ = I₁ × R₁ = 0.0764 × 41 = 3.13 V.
V₂ = V₁ = 3.13 V (at balance, the voltage at both midpoints is the same, so V across the bridge is zero).
I₂ = V₂ / R₂ = 3.13 / 122 = 0.0256 A.
R_x = R₂ × R₃ / R₁ = 122 × 116 / 41 = 345.17 Ω. (The homework writes R_x = R₂R₃/R₁.)
The Wheatstone bridge balance condition can be written several equivalent ways:
R₁ × R_x = R₂ × R₃ (products of opposite arms are equal)
R₁/R₂ = R₃/R_x (ratios of adjacent arms are equal)
V_bridge = 0 (no potential difference across the detector)
This is worth memorising. It comes up in both exam problems and lab work.
Relationship | Formula | When to use |
|---|---|---|
KCL | ΣI_in = ΣI_out at any node | Every multi-loop problem |
KVL | ΣV = 0 around any closed loop | Every multi-loop problem |
Wheatstone balance | R₁ × R_x = R₂ × R₃ | Bridge is balanced (I_bridge = 0) |
Internal resistance | V_terminal = EMF − Ir | Real batteries |
Current divider | I₁ = I_total × R₂/(R₁ + R₂) | Parallel split, two branches |
Power dissipation | P = I²R = V²/R = IV | Finding heat/energy in a resistor |
The Wheatstone bridge is the sensing element in strain gauges, temperature sensors (RTDs), and precision measurement instruments. When the unknown resistance changes (due to strain, temperature, light, etc.), the bridge goes out of balance, and the imbalance voltage is proportional to the change. This is how bathroom scales, load cells in industrial equipment, and many biomedical sensors work.
Multi-loop analysis with Kirchhoff's laws is how circuit simulation software (SPICE and its descendants) solves every circuit you will ever design or debug in practice.
"I can always reduce a multi-loop circuit to a single equivalent resistance." You can do this only when there is a single source. With two or more independent sources, you need Kirchhoff's laws (or superposition, or Thevenin/Norton equivalents for each source).
"A negative current means I made an error." A negative current means the actual current flows opposite to the direction you assumed. The magnitude is still correct. Do not flip signs and re-solve.
"The Wheatstone bridge balance condition only works for identical resistors." It works for any four resistances. The condition R₁R_x = R₂R₃ is general. Equal resistors are a special case, not a requirement.
"Internal resistance can be ignored." In many textbook problems it is set to zero, but when a problem gives you both EMF and terminal voltage, internal resistance is the reason they differ, and you need it.
⚠️ Multi-loop Kirchhoff problems are nearly guaranteed on exams. The algebra is where students lose marks, not the physics. Write your equations clearly, label every current, and check that the number of independent equations matches the number of unknowns before you start solving.
⚠️ Sign errors in KVL are the single most common mistake. Pick a traversal direction, apply the sign convention consistently, and do not change convention mid-loop.
⚠️ Wheatstone bridge problems often ask you to find R_x. Memorise R₁R_x = R₂R₃ and know which resistors are "opposite" in the diamond.
⚠️ Problems involving internal resistance typically ask for r, or for the terminal voltage, or for the power delivered to the external circuit vs. total power. Know V_terminal = EMF − Ir.
True or False: If you assume a current flows clockwise and the answer comes out negative, the current flows anticlockwise.
Fill in the blank: At a balanced Wheatstone bridge, the current through the galvanometer is ________.
True or False: KVL states that the sum of currents at a node is zero.
Fill in the blank: The terminal voltage of a battery with EMF = 12 V and internal resistance r = 2 Ω delivering 3 A is ________ V.
True or False: In a current divider with two parallel resistors, the larger resistor carries the larger current.
Answers: 1. True. 2. Zero. 3. False (that is KCL; KVL is about voltages around a loop). 4. 6 V (V = 12 − 3 × 2). 5. False (the smaller resistor carries more current).
Q: A circuit has two voltage sources, V₁ = 18 V and V₂ = 12 V, with several resistors forming two loops. You write KCL and KVL and arrive at two equations in two unknowns. What is the next step?
A: Solve the system of simultaneous equations by substitution or elimination (or matrix methods). Once you have the branch currents, use Ohm's Law to find any voltages or powers the problem asks for.
Q: A Wheatstone bridge has R₁ = 41 Ω, R₃ = 116 Ω, R₂ = 122 Ω, and is balanced. What is the unknown resistance R_x?
A: R_x = R₂ × R₃ / R₁ = 122 × 116 / 41 ≈ 345 Ω.
Q: A battery has an EMF of 12 V. When it delivers 0.133 A through an external circuit, the terminal voltage is 11.58 V. What is the internal resistance?
A: r = (EMF − V_terminal) / I = (12 − 11.58) / 0.133 ≈ 3.16 Ω.
Q: In a circuit with a single battery and a network of resistors, the total current from the battery is I₁ = 0.133 A. This current splits between R₂ = 82 Ω and a branch with equivalent resistance R₃₄₅ = 208 Ω. What is the current through R₂?
A: By the current divider: I₂ = I₁ × R₃₄₅ / (R₂ + R₃₄₅) = 0.133 × 208 / (82 + 208) = 0.133 × 208 / 290 ≈ 0.095 A.
Q: What is the power dissipated in R₂ = 82 Ω if the current through it is 0.095 A?
A: P = I² × R = (0.095)² × 82 ≈ 0.74 W.
Kirchhoff's laws are the foundation for every circuit analysis technique you will meet: superposition, Thevenin and Norton equivalents, mesh analysis, and nodal analysis all build directly on KVL and KCL. The Wheatstone bridge connects to sensor physics and instrumentation courses. Internal resistance links to the concept of maximum power transfer (which occurs when R_load = r), a result you will see in AC circuits as impedance matching.
Kirchhoff's current law, Kirchhoff's voltage law, KCL, KVL, multi-loop circuit, simultaneous equations circuit, Wheatstone bridge, bridge balance condition, unknown resistance, internal resistance battery, terminal voltage, current divider, voltage divider, power dissipation resistor, PHYS 212 UIUC, University Physics Electricity and Magnetism, mesh analysis, loop analysis