use the formula for the combined resistance of two or more resistors in series

Kirchhoff’s Laws and Resistive Circuits (Cambridge 9702 – Topic 10.2)

These notes cover everything required for the AS‑Level syllabus on Kirchhoff’s laws, series and parallel resistors, mixed circuits, potential dividers and the effect of internal resistance. Each section contains a concise derivation, a worked example, a quick‑check summary and, where appropriate, an alternative method.

1. Kirchhoff’s Laws

1.1 Kirchhoff’s Current Law (KCL)

• Statement: The algebraic sum of currents at a node is zero.

  \$\sum I{\text{in}}-\sum I{\text{out}}=0\qquad\text{or}\qquad\sum I{\text{in}}=\sum I{\text{out}}\$

• Origin – conservation of charge: No net charge can accumulate at a junction; whatever charge flows in must flow out at the same instant.
• Sign convention: Currents entering the node are taken as positive, those leaving as negative (or vice‑versa – be consistent).
• Diagram (node with three branches):

  

1.2 Kirchhoff’s Voltage Law (KVL)

• Statement: The algebraic sum of potential differences (voltage rises + drops) round any closed loop is zero.

  \$\sum V{\text{rise}}-\sum V{\text{drop}}=0\qquad\text{or}\qquad\sum V=0\$

• Origin – conservation of energy: A test charge that travels once around a closed loop returns to its original potential energy; the work done by sources equals the work done against resistive elements.
• Sign convention:

  • Moving in the direction of a source emf (from – to +) is a rise (+V).
  • Moving through a resistor in the direction of current is a drop (‑IR).

• Diagram (single‑loop series circuit):

  

1.3 Worked Example – Two‑Loop (Mesh) Circuit

Battery \$E1=12\,\$V supplies the left loop, battery \$E2=6\,\$V supplies the right loop. The circuit contains three resistors \$R1=2\;\Omega\$, \$R2=4\;\Omega\$, \$R_3=6\;\Omega\$ as shown below.

1. Apply KVL to the left loop (including the shared \$R_2\$):

   \$E1 - I1R1 - (I1-I2)R2 =0\$

2. Apply KVL to the right loop:

   \$E2 - I2R3 - (I2-I1)R2 =0\$

3. Rewrite as simultaneous equations:

   \[

   \begin{cases}

   (R1+R2)I1 - R2 I2 = E1\\[4pt]

   -R2 I1 + (R2+R3)I2 = E2

   \end{cases}

   \]

4. Insert the values and solve:

   \[

   \begin{cases}

   (2+4)I1 - 4 I2 = 12\\

   -4 I1 + (4+6)I2 = 6

   \end{cases}

   \;\Longrightarrow\;

   I1 = 2.0\;\text{A},\quad I2 = 1.0\;\text{A}

   \]

5. Check with KCL at the junction between \$R1\$, \$R2\$, \$R3\$: \$I1 = I2 + (I1-I_2)\$ – satisfied.

Quick‑Check

• When you move with the direction of a current source, write a +V (rise); when you move against it, write ‑V (drop).
• For a resistor, always write ‑IR if you traverse it in the direction of the assumed current.

2. Series Resistances

2.1 Derivation

1. All resistors \$R1,R2,\dots,R_n\$ are end‑to‑end, so the same current \$I\$ flows through each.
2. Apply KVL around the loop:

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

3. Factor \$I\$:

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

4. Define the equivalent (combined) resistance:

   \$\boxed{R{\text{eq(series)}} = \sum{i=1}^{n} R_i}\$

2.2 Power‑Check (optional)

For a series circuit the total power supplied equals the sum of the individual powers:

\[

P{\text{total}} = I^2 R{\text{eq}} = \sum{i=1}^{n} I^2 Ri = \sum{i=1}^{n} Pi

\]

2.3 Worked Example – Series

Battery \$E=12\,\$V, resistors \$R1=4\;\Omega\$, \$R2=6\;\Omega\$, \$R_3=10\;\Omega\$.

1. \$R_{\text{eq}} = 4+6+10 = 20\;\Omega\$
2. \$I = \dfrac{E}{R_{\text{eq}}}= \dfrac{12}{20}=0.60\,\$A
3. Voltage drops: \$V1=IR1=2.4\,\$V, \$V2=3.6\,\$V, \$V3=6.0\,\$V (check: \$2.4+3.6+6.0=12\,\$V).
4. Power in each resistor: \$P1=I^2R1=1.44\,\$W, \$P2=2.16\,\$W, \$P3=3.60\,\$W; total \$7.20\,\$W = \$I^2R_{\text{eq}}\$.

3. Parallel Resistances

3.1 Derivation

1. All branches share the same potential difference \$V\$ across them.
2. Apply KCL at the node where the branches meet:

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

3. Ohm’s law for each branch: \$Ii = \dfrac{V}{Ri}\$.
4. Substitute into KCL and cancel the common \$V\$ (non‑zero):

   \$\$\frac{V}{R{\text{eq}}}= \frac{V}{R1}+ \frac{V}{R2}+ \dots +\frac{V}{Rn}

   \;\Longrightarrow\;

   \boxed{\frac{1}{R{\text{eq(par)}}}= \sum{i=1}^{n}\frac{1}{R_i}}\$\$

3.2 Power‑Check (optional)

For a parallel network the total power supplied equals the sum of the powers in each branch:

\[

P{\text{total}} = \frac{V^2}{R{\text{eq}}}= \sum{i=1}^{n}\frac{V^2}{Ri}= \sum{i=1}^{n} Pi

\]

3.3 Worked Example – Parallel

Resistors \$R1=2\;\Omega\$, \$R2=3\;\Omega\$, \$R_3=6\;\Omega\$ across a \$12\,\$V source.

1. Calculate \$R_{\text{eq}}\$:

   \$\frac{1}{R_{\text{eq}}}= \frac{1}{2}+\frac{1}{3}+\frac{1}{6}=0.5+0.333+0.167=1.0\$

   \$\Rightarrow R_{\text{eq}} = 1\;\Omega\$

2. Total current \$I_{\text{total}} = \dfrac{12}{1}=12\,\$A.
3. Branch currents: \$I1=12/2=6\,\$A, \$I2=12/3=4\,\$A, \$I_3=12/6=2\,\$A (check: \$6+4+2=12\,\$A).
4. Power in each branch: \$P1=VI1=72\,\$W, \$P2=48\,\$W, \$P3=24\,\$W; total \$144\,\$W = \$V^2/R_{\text{eq}}\$.

4. Mixed (Series‑Parallel) Circuits

4.1 Reduction Strategy (most exam‑style)

1. Identify simple series or parallel groups.
2. Replace each group by its equivalent resistance.
3. Repeat until the whole circuit is reduced to a single \$R_{\text{eq}}\$.
4. Use Ohm’s law to find the total current, then work backwards to obtain individual currents and voltages.

4.2 Worked Example – Reduction Method

Battery \$12\,\$V, series resistor \$R1=2\;\Omega\$, then a parallel block \$R2=4\;\Omega\$ // \$R3=6\;\Omega\$, finally \$R4=3\;\Omega\$ back to the battery.

1. Parallel combination:

   \$\frac{1}{R{23}}=\frac{1}{4}+\frac{1}{6}=0.4167\;\Rightarrow\;R{23}=2.40\;\Omega\$

2. Series total:

   \$R{\text{eq}} = R1+R{23}+R4 = 2+2.40+3 = 7.40\;\Omega\$

3. Total current:

   \$I_{\text{total}} = \frac{12}{7.40}=1.62\;\text{A}\$

4. Voltage across the parallel block:

   \$V{23}= I{\text{total}}R_{23}=1.62\times2.40=3.89\;\text{V}\$

5. Branch currents:

   \$\$I2 = \frac{V{23}}{R_2}=0.97\;\text{A},\qquad

   I3 = \frac{V{23}}{R_3}=0.65\;\text{A}\$\$

   (Check: \$0.97+0.65=1.62\,\$A, satisfying KCL.)

4.3 Alternative Method – Simultaneous KCL/KVL (no reduction)

Define currents \$I1\$ (through \$R1\$), \$I2\$ (through \$R2\$) and \$I3\$ (through \$R3\$). \$I1\$ then splits: \$I1 = I2+I3\$. Apply KVL to the two loops:

\[

\begin{aligned}

\text{Loop A (left)}:&\; 12 - I1R1 - I2R2 =0\\

\text{Loop B (right)}:&\; 12 - I1R1 - I3R3 =0

\end{aligned}

\]

Insert \$R\$‑values and solve the three equations (KCL + two KVL) to obtain the same currents as above.

4.4 Common Pitfalls (quick‑list)

• Forgetting that the current through a series element is the same as the total current.
• Assuming the voltage across a series group is the same – it is the total voltage that is shared, not each resistor.
• When reducing a parallel block, do not forget to keep the correct node connections; swapping nodes can change the circuit topology.
• Neglecting the sign convention in KVL (rise vs. drop) leads to algebraic errors, especially in multi‑loop problems.

5. Potential Divider

• Two resistors \$R1\$ and \$R2\$ in series across a source \$V_{\text{in}}\$.
• Voltage across \$R_2\$ (the “output”):

  \$\boxed{V{\text{out}} = V{\text{in}}\frac{R2}{R1+R_2}}\$

• Derivation: From KVL, \$V{\text{in}} = I(R1+R2)\$ and \$V{\text{out}} = IR_2\$. Eliminate \$I\$ to obtain the formula.

Worked Example – Potential Divider

Battery \$9\,\$V, \$R1=1\;\Omega\$, \$R2=3\;\Omega\$.

\[

V_{\text{out}} = 9\;\text{V}\times\frac{3}{1+3}=9\times0.75=6.75\;\text{V}

\]

Loading Effect (brief note)

If a measuring device of resistance \$RL\$ is connected across \$R2\$, the effective resistance becomes \$R2^{\prime}= \dfrac{R2RL}{R2+RL}\$, reducing \$V{\text{out}}\$.

6. Internal Resistance of a Source

• e.m.f. (\$\mathcal{E}\$) – the open‑circuit voltage of an ideal source.
• Internal resistance (\$r\$) – a real source can be modelled as \$\mathcal{E}\$ in series with \$r\$.
• Terminal voltage (\$V\$) when a load \$R_{\text{L}}\$ draws current \$I\$:

  \$\mathcal{E} - I r - I R{\text{L}} =0\;\Longrightarrow\; V = I R{\text{L}} = \mathcal{E} - I r\$

Worked Example – Internal Resistance

Cell: \$\mathcal{E}=1.5\,\$V, \$r=0.5\;\Omega\$. Load \$R_{\text{L}}=2.5\;\Omega\$.

1. Total resistance \$= r+R_{\text{L}} = 3.0\;\Omega\$.
2. Current \$I = \dfrac{\mathcal{E}}{r+R_{\text{L}}}= \dfrac{1.5}{3.0}=0.50\,\$A.
3. Terminal voltage \$V = I R_{\text{L}} = 0.50\times2.5 = 1.25\,\$V (or \$V=\mathcal{E}-Ir=1.5-0.5\times0.5\$).
4. Power in the load \$P{\text{L}} = VI = 0.625\,\$W; power lost in the internal resistance \$Pr = I^2 r = 0.125\,\$W.

7. Summary Table

8. Suggested Diagrams (place‑holders for the teacher)

• Node diagram for KCL (three‑branch junction).
• Single‑loop series circuit with battery and three resistors (showing \$I\$ and \$V_i\$).
• Parallel network with a common voltage source.
• Two‑loop (mesh) circuit for the simultaneous‑equation example.
• Mixed series‑parallel network (the example in Section 4).
• Potential divider schematic.
• Battery model showing \$\mathcal{E}\$, \$r\$, and external load.

9. Practice Questions (AS‑Level style)

1. Three resistors \$2\;\Omega\$, \$5\;\Omega\$ and \$8\;\Omega\$ are connected in series across a \$15\,\$V battery. Calculate:

   • the total current,
   • the voltage across each resistor.

2. A \$9\,\$V source has internal resistance \$0.3\;\Omega\$. It supplies a load of \$3\;\Omega\$. Find the terminal voltage, the current through the load and the power dissipated in the load.
3. Four resistors \$1\;\Omega\$, \$3\;\Omega\$, \$4\;\Omega\$ and \$x\;\Omega\$ are connected in parallel across a \$12\,\$V source. If the total current drawn from the source is \$2\,\$A, determine \$x\$.
4. In the mixed circuit of Section 4, calculate the power dissipated in each resistor.
5. Explain why KCL is automatically satisfied for a series circuit, but KVL must be applied to find the current.
6. A potential divider is made from \$R1=2\;\Omega\$ and \$R2=6\;\Omega\$ across a \$10\,\$V supply. A voltmeter of internal resistance \$100\;\Omega\$ is connected across \$R_2\$. Determine the reading on the voltmeter (assume ideal leads).
7. Using the two‑loop circuit of Section 1.3, find the power dissipated in each resistor and verify that the total supplied power equals the sum of the three powers.