use Kirchhoff’s laws to solve simple circuit problems

Objective

To derive, understand and apply Kirchhoff’s Current Law (KCL) and Kirchhoff’s Voltage Law (KVL) for analysing simple DC circuits, including series‑parallel networks, the potential‑divider principle and the effect of internal resistance of a cell. The notes are written to meet the requirements of Cambridge IGCSE 9702 AS & A‑Level Physics (DC‑circuit section).

1. Physical Basis of Kirchhoff’s Laws

• KCL – charge conservation: At any junction the algebraic sum of currents is zero because charge cannot accumulate at a node.
  

  \[

  \sum{k=1}^{n} Ik =0\qquad\text{(in = out)}

  \]

• KVL – energy conservation: The net change in electric potential around a closed conducting loop is zero; the work done by sources equals the work done on resistors.
  

  \[

  \sum{k=1}^{m} Vk =0

  \]

2. Circuit‑Diagram Essentials (Syllabus 10.1)

2.1 Standard symbols

2.2 Checklist for a tidy circuit diagram

1. Draw a clear rectangular outline for the whole circuit.
2. Place the source(s) on the left‑hand side; label EMF 𝓔 and internal resistance r.
3. Place resistors, ammeters and voltmeters with their symbols; keep components in a straight line where possible.
4. Label every node (A, B, C…) and every branch current (e.g. I₁, I₂).
5. Indicate the direction you assume for each current (any direction is acceptable).
6. Use consistent units (Ω, V, A) and write values close to the symbols.

3. Fundamental Relations

3.1 Series resistance

When resistors are end‑to‑end the same current I flows through each.

\[

\mathcal{E}=I(R1+R2+\dots)=I\,R{\text{eq}}\quad\Longrightarrow\quad R{\text{eq}}=R1+R2+\dots

\]

3.2 Parallel resistance

All branches share the same voltage V.

\[

I=\frac{V}{R1}+\frac{V}{R2}+\dots=V\!\left(\frac1{R1}+\frac1{R2}+\dots\right)

\]

\[

\Longrightarrow\qquad\frac1{R{\text{eq}}}= \frac1{R1}+\frac1{R_2}+\dots

\]

3.3 EMF, internal resistance and terminal voltage

\[

V{\text{t}}=\mathcal{E}-Ir\qquad(\text{ideal source: }r=0\Rightarrow V{\text{t}}=\mathcal{E})

\]

3.4 Potential‑divider (two resistors in series)

\[

V{R2}=V{\text{total}}\;\frac{R2}{R1+R2},\qquad

V{R1}=V{\text{total}}\;\frac{R1}{R1+R2}

\]

4. Systematic Use of Kirchhoff’s Laws (Syllabus 10.2)

1. Draw & label the circuit (see Section 2).
2. Identify independent loops and nodes.
3. Assign a reference current direction to every branch.
4. Write KCL equations for all nodes except one (the omitted node equation is automatically satisfied).
5. Write KVL equations for each independent loop, using the sign convention:

   • Moving in the direction of a current through a resistor → voltage drop (‑ IR).
   • Moving from – to + across a source → voltage rise (+ 𝓔).

6. Solve the simultaneous linear equations (substitution, elimination or matrix method).
7. Check:

   • All KCL equations hold.
   • All KVL equations hold.
   • The equivalent resistance obtained from the currents matches the shortcut series/parallel result.

5. Worked Examples

5.1 Example A – Shortcut method (series‑parallel)

Problem: A 12 V ideal battery supplies three resistors: \(R1=2\;\Omega\) and \(R2=3\;\Omega\) in parallel, the combination in series with \(R_3=4\;\Omega\). Find the current through each resistor.

1. Parallel equivalent:

   \[

   \frac1{R{12}}=\frac1{2}+\frac1{3}=\frac56\;\Longrightarrow\;R{12}=1.20\;\Omega

   \]

2. Total resistance:

   \[

   R{\text{eq}}=R{12}+R_3=1.20+4.00=5.20\;\Omega

   \]

3. Total current:

   \[

   I=\frac{12}{5.20}=2.31\;\text{A}

   \]

4. Voltage across the parallel branch:

   \[

   V{p}=I\,R3=2.31\times4.00=9.23\;\text{V}

   \]

   \[

   V_{\text{parallel}}=12-9.23=2.77\;\text{V}

   \]

5. Currents in the parallel resistors:

   \[

   I1=\frac{V{\text{parallel}}}{R_1}= \frac{2.77}{2}=1.38\;\text{A}

   \]

   \[

   I2=\frac{V{\text{parallel}}}{R_2}= \frac{2.77}{3}=0.92\;\text{A}

   \]

   (Check: \(I1+I2=2.31\;\text{A}=I\).)

5.2 Example B – Full Kirchhoff method (same circuit)

Problem: Solve the circuit of Example A using KCL and KVL.

1. Assign currents (see diagram):

   • \(I\) – total current from the battery (through \(R_3\)).
   • \(I1\) – through \(R1\).
   • \(I2\) – through \(R2\).

2. KCL at the node where the parallel branches split:

   \[

   I = I1 + I2 \tag{1}

   \]

3. KVL for the left loop (battery → \(R3\) → \(R1\) → back):

   \[

   12 - I R3 - I1 R1 =0 \;\Longrightarrow\; 12 -4I -2I1 =0 \tag{2}

   \]

4. KVL for the right loop (battery → \(R3\) → \(R2\) → back):

   \[

   12 - I R3 - I2 R2 =0 \;\Longrightarrow\; 12 -4I -3I2 =0 \tag{3}

   \]

5. From (2) and (3):

   \[

   I1 = 6 -2I,\qquad I2 = 4 -\frac{4}{3}I

   \]

6. Insert into (1):

   \[

   I = (6-2I) + \Bigl(4-\frac{4}{3}I\Bigr)

   \;\Longrightarrow\; \frac{13}{3}I =10

   \;\Longrightarrow\; I = 2.31\;\text{A}

   \]

7. Currents in the parallel branches:

   \[

   I_1 = 6-2I = 6-4.62 = 1.38\;\text{A}

   \]

   \[

   I_2 = 4-\frac{4}{3}I = 4-3.08 = 0.92\;\text{A}

   \]

8. All KCL and KVL equations are satisfied; results agree with Example A.

5.3 Example C – Solving a Δ (triangular) network

Problem: A 12 V ideal battery is connected to a Δ network of three resistors: \(R1=3\;\Omega\) (AB), \(R2=6\;\Omega\) (BC), \(R3=2\;\Omega\) (CA). Assign clockwise currents \(I1\) (AB), \(I2\) (BC), \(I3\) (CA). Find all three currents using KVL.

1. Choose node A as the reference (ground). The three independent loops are:

   • Loop 1: AB → BC → CA (clockwise).
   • Loop 2: AB → Battery → CA (clockwise).
   • Loop 3: BC → Battery → CA (clockwise).

2. Write KVL for each loop (rise across the battery = +12 V):

   Loop 1 (AB‑BC‑CA):
   -I₁R₁ - I₂R₂ + I₃R₃ = 0                (4)
   Loop 2 (Battery‑AB‑CA):
   +12 - I₁R₁ - I₃R₃ = 0                 (5)
   Loop 3 (Battery‑BC‑CA):
   +12 - I₂R₂ - I₃R₃ = 0                 (6)
   

3. Substitute the resistor values:

   (4)  -3I₁ - 6I₂ + 2I₃ = 0
   (5)   12 - 3I₁ - 2I₃ = 0   → 3I₁ + 2I₃ = 12
   (6)   12 - 6I₂ - 2I₃ = 0   → 6I₂ + 2I₃ = 12
   

4. Solve (5) for \(I₁\):

   \[

   I₁ = \frac{12-2I₃}{3}

   \]

   Solve (6) for \(I₂\):

   \[

   I₂ = \frac{12-2I₃}{6}= \frac{12-2I₃}{6}

   \]

5. Insert \(I₁\) and \(I₂\) into (4):

   \[

   -3\!\left(\frac{12-2I₃}{3}\right) -6\!\left(\frac{12-2I₃}{6}\right) +2I₃ =0

   \]

   Simplifies to:

   \[

   -(12-2I₃) -(12-2I₃) +2I₃ =0\;\Longrightarrow\; -24+4I₃+2I₃=0

   \]

   \[

   6I₃ = 24\;\Longrightarrow\; I₃ = 4\;\text{A}

   \]

6. Back‑substitute:

   \[

   I₁ = \frac{12-2(4)}{3}= \frac{4}{3}=1.33\;\text{A}

   \]

   \[

   I₂ = \frac{12-2(4)}{6}= \frac{2}{3}=0.67\;\text{A}

   \]

7. Result (clockwise direction assumed):

   • \(I1 = 1.33\;\text{A}\) through \(R1\) (AB)
   • \(I2 = 0.67\;\text{A}\) through \(R2\) (BC)
   • \(I3 = 4.00\;\text{A}\) through \(R3\) (CA)

   Negative values would indicate opposite direction; here all are positive, so the assumed directions are correct.

6. Symbol Table (Expanded)

7. Practice Questions (with marks)

1. Series‑parallel with internal resistance (5 marks)
   

   A 9 V battery has internal resistance \(r=1\;\Omega\). It supplies two external resistors \(R1=5\;\Omega\) and \(R2=10\;\Omega\) in series, and a third resistor \(R3=15\;\Omega\) is placed in parallel with \(R2\). Determine:

   • The current through each resistor.
   • The terminal voltage of the battery.

2. Potential divider with a non‑ideal source (4 marks)
   

   A 6 V battery (internal resistance \(r=0.5\;\Omega\)) drives a series pair \(RA=2\;\Omega\) and \(RB=4\;\Omega\). This series pair is in parallel with \(R_C=6\;\Omega\). Find the voltage across each resistor.

3. Δ‑network – write the equations (3 marks)
   

   For the Δ circuit of Example C, write the three independent KVL equations (do not solve). Indicate clearly the sign you assign to each voltage drop.

4. Potential‑divider verification (2 marks)
   

   In a series circuit \(R1=8\;\Omega\), \(R2=12\;\Omega\) are connected across a 15 V ideal battery. Calculate the voltage across \(R_2\) using the potential‑divider formula and verify by a direct KVL sum.

8. Summary

• KCL – charge cannot build up at a node: \(\displaystyle\sum I{\text{in}}=\sum I{\text{out}}\).
• KVL – energy is conserved round any closed loop: \(\displaystyle\sum V{\text{rise}}=\sum V{\text{drop}}\).
• Series: \(R{\text{eq}}=R1+R2+\dots\). Parallel: \(\displaystyle\frac1{R{\text{eq}}}= \frac1{R1}+ \frac1{R2}+ \dots\).
• EMF \(\mathcal{E}\) is the open‑circuit voltage; the terminal voltage falls by \(Ir\) because of internal resistance.
• Potential‑divider: \(V{R2}=V{\text{total}}\dfrac{R2}{R1+R2}\) – useful for voltage references.
• Systematic Kirchhoff analysis (nodes + loops) always yields a set of linear equations; shortcut series/parallel formulas give a quick check.
• Assumed current directions are arbitrary – a negative answer simply means the actual direction is opposite.
• Good circuit diagrams (Section 2) and a clear checklist help avoid algebraic mistakes.