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

Published by Patrick Mutisya · 14 days ago

Cambridge A-Level Physics 9702 – Kirchhoff’s Laws

Kirchhoff’s Laws

Kirchhoff’s laws are fundamental tools for analysing complex electric circuits. They allow us to relate currents and voltages in any network of resistors, batteries and other components.

1. Kirchhoff’s Current Law (KCL)

KCL states that the algebraic sum of currents meeting at a node (junction) is zero:

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

In words, the total current entering a junction equals the total current leaving it.

2. Kirchhoff’s \cdot oltage Law (K \cdot L)

K \cdot L states that the algebraic sum of the potential differences (voltages) around any closed loop is zero:

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

This reflects the conservation of energy for a charge moving around the loop.

3. Series Resistors and Combined Resistance

When resistors are connected end‑to‑end, they form a series circuit. The same current flows through each resistor, and the total voltage across the series combination is the sum of the individual voltage drops.

The combined (equivalent) resistance \$R{\text{eq}}\$ of \$n\$ resistors \$R1, R2, \dots , Rn\$ in series is given by:

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

4. Using K \cdot L with Series Resistors

Consider a simple series loop containing a battery of emf \$E\$ and three resistors \$R1\$, \$R2\$, \$R_3\$. Applying K \cdot L:

\$E - I R1 - I R2 - I R_3 = 0\$

Factorising the current \$I\$ gives:

\$E = I (R1 + R2 + R3) = I R{\text{eq}}\$

Hence the current in the loop is:

\$I = \frac{E}{R_{\text{eq}}}\$

5. Example Calculation

  1. Given: \$E = 12\ \text{V}\$, \$R1 = 4\ \Omega\$, \$R2 = 6\ \Omega\$, \$R_3 = 10\ \Omega\$.

  2. Find the equivalent resistance:

    \$R_{\text{eq}} = 4\ \Omega + 6\ \Omega + 10\ \Omega = 20\ \Omega\$

  3. Determine the current using K \cdot L:

    \$I = \frac{12\ \text{V}}{20\ \Omega} = 0.60\ \text{A}\$

  4. Voltage drop across each resistor:

    • \$V1 = I R1 = 0.60\ \text{A} \times 4\ \Omega = 2.4\ \text{V}\$

    • \$V2 = I R2 = 0.60\ \text{A} \times 6\ \Omega = 3.6\ \text{V}\$

    • \$V3 = I R3 = 0.60\ \text{A} \times 10\ \Omega = 6.0\ \text{V}\$

  5. Check K \cdot L: \$2.4\ \text{V} + 3.6\ \text{V} + 6.0\ \text{V} = 12\ \text{V}\$ (satisfied).

6. Summary Table

ConceptMathematical FormKey Point for Series Circuits
KCL\$\displaystyle \sum I{\text{in}} = \sum I{\text{out}}\$Current is the same through all series elements.
K \cdot L\$\displaystyle \sum V{\text{rise}} = \sum V{\text{drop}}\$Sum of voltage drops equals the source emf.
Series Resistance\$\displaystyle R{\text{eq}} = \sum{i=1}^{n} R_i\$Resistances add directly.
Current in Series Loop\$\displaystyle I = \dfrac{E}{R_{\text{eq}}}\$Same current flows through each resistor.

7. Suggested Diagram

Suggested diagram: A single loop containing a battery (emf \$E\$) and three series resistors \$R1\$, \$R2\$, \$R3\$ labelled with the direction of current \$I\$ and voltage drops \$V1\$, \$V2\$, \$V3\$.

8. Practice Questions

  1. Three resistors \$2\ \Omega\$, \$5\ \Omega\$ and \$8\ \Omega\$ are connected in series across a \$15\ \text{V}\$ battery. Calculate the current and the voltage across each resistor.

  2. A circuit contains a \$9\ \text{V}\$ source and four series resistors with values \$1\ \Omega\$, \$3\ \Omega\$, \$4\ \Omega\$, and \$x\ \Omega\$. If the total current is \$0.5\ \text{A}\$, find \$x\$.

  3. Explain why KCL is automatically satisfied for a series circuit, but K \cdot L must be applied to determine the current.