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Kirchhoff’s Laws: KCL and KVL with Solved Examples
Learn Kirchhoff’s Current Law (KCL) and Kirchhoff’s Voltage Law (KVL) in simple words: statements, equations, sign convention, a step-by-step method for solving circuits, limitations and 5 solved problems for BEE and GATE EE.
⚡ In 30 seconds
- KCL: at any node, current in = current out, so \(\sum I = 0\). It comes from conservation of charge.
- KVL: around any closed loop, the sum of voltages is zero, \(\sum V = 0\). It comes from conservation of energy.
- Assume current directions first. A negative answer only means the real current flows the other way.
- A network with \(b\) branches and \(n\) nodes needs \(n-1\) KCL and \(b-n+1\) KVL equations.
What Are Kirchhoff’s Laws?
For a single resistor, Ohm’s law (\(V = IR\)) is enough. But real circuits have many branches, several sources and more than one loop. Ohm’s law alone cannot tell us how the current divides between the branches. For such circuits we use the two laws given by the German physicist Gustav Kirchhoff in 1845.
Kirchhoff’s Current Law (KCL)
Deals with currents at a node. Based on conservation of electric charge.
Kirchhoff’s Voltage Law (KVL)
Deals with voltages around a closed loop. Based on conservation of energy.
Both laws work for DC circuits and for AC circuits (using instantaneous values or phasors). Together they are the base of nodal analysis, mesh analysis and every network theorem you will study later.
Circuit Terms: Node, Branch, Loop and Mesh
Before applying the laws, you must know these words. We will use the circuit in Figure 1 to explain them (we will also solve it in Q3).
| Term | Meaning | In Figure 1 |
|---|---|---|
| Node | A point where two or more elements are joined | A and B are the main nodes |
| Junction (principal node) | A node where three or more branches meet | A and B, so \(n = 2\) |
| Branch | A path between two junctions that carries one current | E₁–R₁, R₃, E₂–R₂, so \(b = 3\) |
| Loop | Any closed path that does not pass through a node twice | 3 loops (left, right, outer) |
| Mesh | A loop that has no other loop inside it | 2 meshes (Loop 1 and Loop 2) |
🎯 GATE tip: how many equations?
For a connected network with \(b\) branches and \(n\) nodes, the number of independent KCL equations is \(n-1\), and the number of independent KVL equations is \(b-n+1\). For Figure 1: \(n-1 = 1\) KCL equation and \(3-2+1 = 2\) KVL equations, total 3 equations for 3 unknown currents.
Kirchhoff’s Current Law (KCL)
📖 Statement of KCL
The algebraic sum of currents meeting at a node is zero. In other words, the total current entering a node is equal to the total current leaving it.
∑ KCL
\[ \begin{gathered} \sum I_{\text{in}} = \sum I_{\text{out}} \\ \text{or} \quad \sum_{k} I_k = 0 \end{gathered} \]Explanation with an example node
In Figure 2, currents \(I_1\) and \(I_4\) flow into node O, while \(I_2\) and \(I_3\) flow out of it. By KCL:
\[ I_1 + I_4 = I_2 + I_3 \]Taking entering currents as positive and leaving currents as negative, the same equation becomes
\[ I_1 - I_2 - I_3 + I_4 = 0 \]Physical meaning of KCL
Electric charge can neither be created nor destroyed. A node is just a meeting point of wires, so it cannot store charge. Therefore the charge flowing into the node every second must flow out in the same second. This is why KCL is called the law of conservation of charge for circuits. Think of water pipes: the water entering a pipe joint must leave it, because the joint cannot hold water.
📝 KCL for a closed surface
KCL also holds for any closed boundary drawn around a part of a circuit (a supernode). The total current entering the boundary equals the total current leaving it. This idea is used a lot in nodal analysis.
⚠️ In AC circuits
KCL holds for instantaneous values and for phasors, but not for RMS magnitudes. If 3 A flows in a resistor and 4 A in a capacitor connected in parallel, the supply current is \(\sqrt{3^2 + 4^2} = 5\ \text{A}\), not 7 A (see Q5).
Kirchhoff’s Voltage Law (KVL)
📖 Statement of KVL
In any closed loop of a network, the algebraic sum of all the voltages is zero. Equivalently, the sum of the EMFs in the loop equals the sum of the voltage drops (\(IR\)) in that loop.
∑ KVL
\[ \begin{gathered} \sum V = 0 \\ \text{or} \quad \sum E = \sum IR \end{gathered} \]Explanation with a single loop
Start at the negative terminal of the source and move clockwise around the loop. Crossing the source from − to + we gain \(V\). Crossing each resistor in the direction of the current we lose \(IR\). We come back to the same point, so the total change is zero:
\[ V - IR_1 - IR_2 = 0 \quad\Rightarrow\quad V = IR_1 + IR_2 \]So the voltage supplied by the source is shared by the two resistors.
Physical meaning of KVL
Potential at a point has only one value. If a charge goes around a closed loop and comes back to its starting point, it is back at the same potential, so the net change in its energy is zero. The energy it gains from the sources is fully used up in the resistors and other elements. This is why KVL is called the law of conservation of energy for circuits. Think of walking around a hill and returning to the same spot: however much you climbed, you came down by exactly the same height.
Sign Convention for KVL
Most mistakes in KVL come from signs. Choose a direction of travel around the loop (clockwise is common) and apply these four rules:
| Element | Direction of travel | Term in KVL |
|---|---|---|
| Resistor | Same direction as the current | \(-IR\) (fall of potential) |
| Resistor | Opposite to the current | \(+IR\) (rise of potential) |
| Battery / source | From − terminal to + terminal | \(+E\) (rise of potential) |
| Battery / source | From + terminal to − terminal | \(-E\) (fall of potential) |
💡 Remember
The sign of a battery depends only on which terminal you meet first, not on the direction of the current. The sign of an \(IR\) term depends on the direction of the current. Any consistent convention gives the same final answer.
Steps to Solve a Circuit Using KCL and KVL
- Mark the unknown currents. Give every branch a current and assume a direction. Do not worry about guessing wrong.
- Apply KCL at \(n-1\) junctions to relate the branch currents.
- Apply KVL to \(b-n+1\) independent loops (the meshes are the easiest choice), using the sign convention above.
- Solve the equations together (substitution, elimination or Cramer’s rule).
- Read the signs. A negative current means it actually flows opposite to your assumed direction; its size is still correct.
- Check with KCL at another node or with a power balance (power delivered = power absorbed).
Difference Between KCL and KVL
| Point | KCL | KVL |
|---|---|---|
| Deals with | Currents | Voltages |
| Applied at | A node or junction | A closed loop or mesh |
| Based on | Conservation of charge | Conservation of energy |
| Equation | \(\sum I = 0\) | \(\sum V = 0\) |
| Independent equations | \(n-1\) | \(b-n+1\) |
| Main method | Nodal analysis | Mesh analysis |
Limitations of Kirchhoff’s Laws
Kirchhoff’s laws assume a lumped circuit: every element is very small compared with the wavelength of the signal, and fields stay inside the elements. Because of this:
- They are not accurate at very high frequencies (long transmission lines, antennas), where the circuit size is comparable to the wavelength.
- KVL in its simple form fails if a changing magnetic flux links the loop from outside. The induced EMF must then be added to the loop as a source.
- KCL in its simple form fails at a point where charge can build up, such as one plate of a capacitor. Treat the whole capacitor as one element and KCL holds again.
- In AC circuits, currents and voltages must be added as phasors, not as plain RMS numbers.
Solved Problems on KCL and KVL
Try each question first, then tap Show solution.
In Figure 2, \(I_1 = 5\ \text{A}\), \(I_4 = 3\ \text{A}\) and \(I_2 = 2\ \text{A}\). Find \(I_3\).
Show solution
Current in = current out at node O:
\[ \begin{aligned} I_1 + I_4 &= I_2 + I_3 \\ 5 + 3 &= 2 + I_3 \\ I_3 &= 6\ \text{A} \end{aligned} \]A 12 V battery and a 4 V battery are connected in series opposing each other (+ terminal facing + terminal) with resistors \(R_1 = 2\ \Omega\) and \(R_2 = 6\ \Omega\) in a single loop. Find the current and the voltage across each resistor.
Show solution
The 12 V battery is larger, so assume the current flows in its direction. Going round the loop along the current, we cross the 12 V battery from − to + (\(+12\)), the 4 V battery from + to − (\(-4\)), and both resistors along the current:
\[ \begin{aligned} 12 - 4 - 2I - 6I &= 0 \\ 8I &= 8 \\ I &= 1\ \text{A} \end{aligned} \] \[ \begin{aligned} V_{R1} &= IR_1 = 1 \times 2 = 2\ \text{V} \\ V_{R2} &= IR_2 = 1 \times 6 = 6\ \text{V} \end{aligned} \]Check: \(2 + 6 = 8\ \text{V} = 12 - 4\). ✅
For the circuit in Figure 1, find the branch currents \(I_1\), \(I_2\) and \(I_3\), and check the power balance.
Show solution
Step 1: KCL at node A
\[ I_3 = I_1 + I_2 \]Step 2: KVL in Loop 1 (E₁, R₁, R₃)
\[ \begin{aligned} 12 - 2I_1 - 6I_3 &= 0 \\ 12 - 2I_1 - 6(I_1 + I_2) &= 0 \\ 8I_1 + 6I_2 &= 12 \end{aligned} \]Step 3: KVL in Loop 2 (E₂, R₂, R₃)
\[ \begin{aligned} 6 - 3I_2 - 6I_3 &= 0 \\ 6 - 3I_2 - 6(I_1 + I_2) &= 0 \\ 6I_1 + 9I_2 &= 6 \end{aligned} \]Step 4: Solve
Multiply the Loop 1 equation by 3 and the Loop 2 equation by 2, then subtract:
\[ \begin{aligned} 24I_1 + 18I_2 &= 36 \\ 12I_1 + 18I_2 &= 12 \end{aligned} \]Subtracting the second from the first:
\[ 12I_1 = 24 \;\Rightarrow\; I_1 = 2\ \text{A} \] \[ \begin{aligned} 6I_2 &= 12 - 8I_1 = 12 - 16 = -4 \\ I_2 &= -\tfrac{2}{3}\ \text{A} \approx -0.667\ \text{A} \\ I_3 &= I_1 + I_2 = 2 - \tfrac{2}{3} \\ &= \tfrac{4}{3}\ \text{A} \approx 1.333\ \text{A} \end{aligned} \]\(I_2\) is negative, so the real current in the right branch flows away from A, into the + terminal of E₂. The 12 V source is actually charging the 6 V source.
Step 5: Power balance
E₁ delivers power. E₂ (being charged) and the three resistors absorb power.
\[ \begin{aligned} P_{E1} &= 12 \times 2 = 24\ \text{W} \\ P_{E2} &= 6 \times \tfrac{2}{3} = 4\ \text{W} \\ P_{R1} &= 2^2 \times 2 = 8\ \text{W} \\ P_{R2} &= \left(\tfrac{2}{3}\right)^2 \times 3 = \tfrac{4}{3}\ \text{W} \\ P_{R3} &= \left(\tfrac{4}{3}\right)^2 \times 6 = \tfrac{32}{3}\ \text{W} \end{aligned} \] \[ 4 + 8 + \tfrac{4}{3} + \tfrac{32}{3} = 24\ \text{W} \;\;\checkmark \]A connected network has 7 branches and 4 nodes. How many independent KCL and KVL equations can be written?
Show solution
Total \(3 + 4 = 7\), equal to the number of unknown branch currents. ✅
A resistor and a capacitor are connected in parallel across an AC supply. An ammeter reads 3 A in the resistor branch and 4 A in the capacitor branch. What does an ammeter in the supply line read?
Show solution
The resistor current is in phase with the voltage, while the capacitor current leads the voltage by \(90^\circ\). KCL must be applied to phasors:
\[ \begin{aligned} \bar I &= \bar I_R + \bar I_C = 3 + j4\ \text{A} \\ |\bar I| &= \sqrt{3^2 + 4^2} = \sqrt{25} = 5\ \text{A} \end{aligned} \]Simply adding the meter readings (3 + 4 = 7 A) is wrong.
Want more practice? See our KVL solved example problems.
Common Mistakes and Quick Summary
- Mixing sign conventions inside the same loop equation.
- Changing the assumed current direction halfway through the problem.
- Forgetting a voltage drop, or counting the same element twice in a loop.
- Writing more KVL equations than \(b-n+1\); the extra ones are not independent.
- Adding AC currents or voltages as plain numbers instead of phasors.
∑ Remember
\[ \begin{aligned} \text{KCL:}&\quad \sum I_{\text{in}} = \sum I_{\text{out}} \\ \text{KVL:}&\quad \sum E = \sum IR \end{aligned} \]Apply KCL at a node and KVL around a closed loop. KCL is the base of nodal analysis; KVL is the base of mesh analysis.
Frequently Asked Questions
What is Kirchhoff’s Current Law (KCL)?
KCL states that the algebraic sum of currents meeting at a node is zero. The total current entering a node equals the total current leaving it. It is based on conservation of charge.
What is Kirchhoff’s Voltage Law (KVL)?
KVL states that the algebraic sum of all voltages around any closed loop is zero, so the sum of EMFs equals the sum of IR drops in the loop. It is based on conservation of energy.
Are Kirchhoff’s laws valid for AC circuits?
Yes, for instantaneous values and for phasors. RMS readings cannot be added directly; they must be added as phasors.
What does a negative current mean in KCL or KVL problems?
It means the actual current flows opposite to the direction you assumed. Its magnitude is still correct, so you do not need to solve again.
How many KVL equations are needed for a circuit?
For a connected network with b branches and n nodes, b − n + 1 independent KVL equations and n − 1 independent KCL equations are needed.
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📌 Further reading & related tutorials
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