recall Kirchhoff’s first law and understand that it is a consequence of conservation of charge

Kirchhoff’s Laws – Cambridge IGCSE / A‑Level Physics (9702)

Learning Objectives

• Recall and state Kirchhoff’s 1st law (Current Law) and explain why it follows from the conservation of electric charge.
• Recall and state Kirchhoff’s 2nd law (Voltage Law) and explain its link to energy conservation.
• Derive the series‑ and parallel‑resistance formulas using the two laws together with Ohm’s law.
• Apply KCL, KVL and Ohm’s law to solve DC‑circuit problems, including:

  • potential‑divider networks,
  • internal resistance of a source,
  • current‑division in parallel branches, and
  • combined series‑parallel resistor arrangements.

• Connect the laws to later topics such as electric fields, power, and energy.

1. Kirchhoff’s First Law – Current Law (KCL)

Statement – At any node (junction) in a steady‑state circuit the algebraic sum of currents is zero:

\$\displaystyle\sum{k=1}^{n} Ik = 0\$

Sign convention

• Choose a direction for each current at the node.
• If the chosen direction is toward the node, treat the current as positive.
• If the chosen direction is away from the node, treat it as negative.
• If the real current flows opposite to the assumed direction, the solved value will be negative – the law still holds.

Diagram (node example)

Why the law holds – Conservation of charge

In a short time \$\Delta t\$ the charge entering the node is \$Q{\text{in}}=\sum I{\text{in}}\Delta t\$, and the charge leaving is \$Q{\text{out}}=\sum I{\text{out}}\Delta t\$.

Because a node cannot store net charge in the steady state, \$Q{\text{in}}=Q{\text{out}}\$, giving

\$\sum I{\text{in}}-\sum I{\text{out}}=0\;\Longrightarrow\;\sum{k=1}^{n}Ik=0.\$

2. Kirchhoff’s Second Law – Voltage Law (KVL)

Statement – Traversing any closed loop, the algebraic sum of potential differences is zero:

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

Sign convention (loop direction)

• Choose a direction around the loop (clockwise or anticlockwise).
• When you move with the current through a resistor, the potential change is a drop: \$-IR\$.
• When you move against the current through a resistor, the change is a rise: \$+IR\$.
• Moving from the – to the + terminal of a battery (or emf source) is a rise: \$+E\$; the opposite direction is a drop: \$-E\$.

Limitation (syllabus wording)

KVL is valid for circuits in a steady state where the magnetic flux through any loop is constant. It does not apply to circuits containing inductors with a time‑varying current unless the induced emf is explicitly included in the loop equation.

Diagram (simple loop)

3. Deriving the Resistance Formulas

3.1 Series resistance

1. All resistors share the same current \$I\$ (KCL).
2. Applying KVL around the series loop:

   \$E - I R1 - I R2 - \dots - I R_n = 0\$

   \$\Rightarrow\; I\,(R1+R2+\dots+R_n)=E\$

3. Therefore the equivalent resistance is

   \$\boxed{R{\text{eq(series)}} = R1+R2+\dots+Rn}\$

3.2 Parallel resistance

1. All resistors are connected across the same two nodes, so they have the same voltage \$V\$ (KVL).
2. Ohm’s law for each branch:

   \$\boxed{Ik = \dfrac{V}{Rk}} \qquad (k=1,2,\dots,n)\$

3. KCL at the top node gives the total current:

   \$\$I{\text{total}} = I1+I2+\dots+In

   = V\!\left(\frac1{R1}+\frac1{R2}+\dots+\frac1{R_n}\right)\$\$

4. Define \$R{\text{eq}}\$ by \$V = I{\text{total}}R_{\text{eq}}\$, yielding

   \$\boxed{\displaystyle\frac1{R{\text{eq(parallel)}}}= \frac1{R1}+ \frac1{R2}+ \dots+ \frac1{Rn}}.\$

3.3 Current‑division rule (used for parallel branches)

For a parallel network fed by a total current \$I_{\text{total}}\$, the current in branch \$i\$ is

\$\boxed{Ii = I{\text{total}}\;\frac{R{\text{eq}}}{Ri}}\$

where \$R_{\text{eq}}\$ is the equivalent resistance of the whole parallel group (the denominator of the parallel‑resistance formula). This form is explicitly required by the Cambridge syllabus.

4. Potential Divider (Topic 10.3)

Two resistors \$R1\$ and \$R2\$ in series across a battery of emf \$E\$ give a voltage across \$R_2\$ of

\$V{R2}=E\;\frac{R2}{R1+R_2}.\$

Numerical example (exam style)

• \$E=12\;\text{V},\;R1=2\;\Omega,\;R2=8\;\Omega\$
• \$V{R2}=12\;\text{V}\times\dfrac{8}{2+8}=9.6\;\text{V}\$
• Current in the series string: \$I=E/(R1+R2)=12/10=1.2\;\text{A}\$.

5. Internal Resistance of a Source

A real battery is modelled as an ideal emf \$E\$ in series with an internal resistance \$r\$.

• Terminal voltage (voltage across the external load) is

  \$\boxed{V_{\text{term}} = E - I r}\$

• Applying KVL to the whole loop gives the current drawn by a load \$R_{\!L}\$:

  \$\$E - I r - I R_{\!L}=0\;\Longrightarrow\;

  \boxed{I = \dfrac{E}{r+R_{\!L}}}\$\$

Practical note (Paper 3) – \$r\$ can be measured by recording the terminal voltage for two different loads, plotting \$V\$ against \$I\$, and finding the slope (which equals \$-r\$).

6. Power and Energy (Link to later topics)

• Power in any element:

  \$P = IV = I^{2}R = \dfrac{V^{2}}{R}\$

• Energy supplied over a time \$t\$:

  \$W = Pt = \int P\,dt\$

• These relations are used in the “Power & Energy” part of the syllabus and in later A‑Level topics on electric fields.

7. Worked Example – Three‑branch Network

Problem statement

• A 12 V battery has internal resistance \$r = 1\;\Omega\$.
• External network:

  • \$R_1 = 4\;\Omega\$ (branch A, series with the parallel part)
  • \$R_2 = 6\;\Omega\$ (branch B)
  • \$R_3 = 12\;\Omega\$ (branch C)

• Branches B and C are in parallel; this parallel combination is in series with \$R_1\$.
• Find:

  1. Total current supplied by the battery.
  2. Current in each resistor.
  3. Terminal voltage of the battery.

Solution steps

1. Combine the parallel part (B‖C) using the parallel formula:

   \$\$\frac1{R_{BC}} = \frac1{6} + \frac1{12}= \frac{2+1}{12}= \frac{3}{12}

   \;\Longrightarrow\; \boxed{R_{BC}=4\;\Omega}\$\$

2. Form the overall series resistance (including internal \$r\$):

   \$R{\text{eq}} = r + R1 + R_{BC}= 1 + 4 + 4 = 9\;\Omega\$

3. Total current (KVL around the whole loop):

   \$I{\text{tot}} = \frac{E}{R{\text{eq}}}= \frac{12\;\text{V}}{9\;\Omega}= \boxed{1.33\;\text{A}}\$

4. Currents in series elements (same as \$I_{\text{tot}}\$):

   \$I{R1}= I_{\text{tot}} = 1.33\;\text{A}\$

5. Current division in the parallel part (use the rule from §3.3):

   Equivalent resistance of the parallel group is \$R_{BC}=4\;\Omega\$.

   \$\$I{R2}= I{\text{tot}}\;\frac{R{BC}}{R_2}

   = 1.33\;\text{A}\times\frac{4}{6}=0.89\;\text{A}\$\$

   \$\$I{R3}= I{\text{tot}}\;\frac{R{BC}}{R_3}

   = 1.33\;\text{A}\times\frac{4}{12}=0.44\;\text{A}\$\$

   (Check: \$I{R2}+I{R3}=1.33\;\text{A}=I_{\text{tot}}\$.)

6. Terminal voltage (voltage across the external network):

   \$\$V{\text{term}} = E - I{\text{tot}}\,r

   = 12\;\text{V} - (1.33\;\text{A})(1\;\Omega)= \boxed{10.7\;\text{V}}\$\$

Result summary

Diagram (network used in the example)

8. Summary Table

9. Links to Later Topics

• Electric fields: The potential differences used in KVL are line integrals of the electric field, \$\Delta V = -\int\mathbf{E}\cdot d\mathbf{l}\$.
• Power & energy: Once \$I\$ and \$V\$ are known, \$P=IV\$ gives the rate of energy transfer; \$W = Pt\$ links to work and energy concepts in later A‑Level modules.
• Practical skills: Measuring internal resistance, verifying KVL with a voltmeter, and checking KCL with an ammeter are common Paper 3 tasks.

10. Key Points to Remember

• KCL is always valid for any node in a steady‑state circuit – it is a direct statement of charge conservation.
• KVL holds for any closed loop provided the magnetic flux through the loop is constant (no induced emf unless included).
• Sign conventions are a matter of choice; a wrong initial guess simply yields a negative answer after solving.
• Series and parallel resistance formulas are derived from the two laws together with Ohm’s law.
• Current‑division and potential‑divider formulas are practical shortcuts that stem from KCL and KVL.
• Power formulas \$P=IV=I^{2}R=V^{2}/R\$ and the energy relation \$W=Pt\$ are essential for later topics.