# Analysis of Complex Electric Circuit Problems

This topic is part of the HSC Physics course under the section Electric Circuits.

### HSC Physics Syllabus

• investigate qualitatively and quantitatively series and parallel circuits to relate the flow of current through the individual components, the potential differences across those components and the rate of energy conversion by the components to the laws of conservation of charge and energy, by deriving the following relationships: (ACSPH038, ACSPH039, ACSPH044)
– \sum I = 0 (Kirchhoff’s current law – conservation of charge)
– \sum V = 0 ((Kirchhoff’s voltage law – conservation of energy)
– R_{\text{Series}} = R_1 + R_2 + \ldots + R_n
– \frac{1}{R_{\text{Parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots + \frac{1}{R_n}

### Complex Circuit Problems in HSC Physics

Navigating the maze of complex electric circuit problems can be daunting for Year 11 HSC Physics students. As you progress in your studies, circuits won't always be simple series or parallel connections; they'll often be a combination of both. In this video, we discuss how to approach these problems using two examples.

### Break It Down and Apply Theories

The first step is to simplify the problem. Break down the complex circuit into identifiable series and parallel sections. Remember, in a series circuit, components are daisy-chained so the current through each component is the same. In parallel circuits, components are like rungs on a ladder, sharing the same two nodes and thus the same voltage.

Once you've identified these simpler sections, it's time to apply

• Ohm's Law: V=IR
• Kirchhoff's Current Law tells us that all current entering a node must equal the current leaving it
• Kirchhoff's Voltage Law states that the sum of all voltages around a loop must equal zero
• Electrical power: P = VI

### Problem 1

Consider the following circuit.

Calculate the current flowing through each resistor.

Solution:

First work out the total resistance of the circuit in order to calculate the total current.

The resistance of the two 4 Ohm resistor connected in parallel is given by:

$$\frac{1}{R} = \frac{1}{4} + \frac{1}{4}$$

$$R = 2 \, \Omega$$

The parallel circuit is connected to the 6 Ohms resistor in series. Thus, the total resistance of the circuit is given by:

$$R = 6 + 2 = 8 \, \Omega$$

The total current is given by Ohm's law:

$$I = \frac{V}{R}$$

$$I = \frac{12}{8} = 1.5 \text{ A}$$

The 6 Ohms resistor is connected in series so all charges must flow through it, giving it a current of 1.5 A.

The two 4 Ohms resistor will receive 1.5 A of current too (conservation of charge), and since they have equal resistance, the current is divided equally between them - each resistor has 0.75 A of current.

### Problem 2

The current flowing through R1 and R2 are measured to be 2 A and 1 A respectively. The resistance of R2 is 5 Ω.

Calculate the resistance of R1 and R3.

Solution:

Since R1 is connected in series, the current flowing through it is also the total current in the circuit (conservation of charge).

Thus the total resistance of the circuit is given by:

$$R_{\text{total}} = \frac{V}{I}$$

$$R_{\text{total}} = \frac{10}{2} = 5 \, \Omega$$

The voltage across R2 is given by:

$$V = 1 \times 5 = 5 \text{ V}$$

This means the voltage across R3 is also 5 V (conservation of energy: voltage is constant in parallel circuit).

If a current of 1 A flows through R2, then 1 A should also flow through R3 (KCL).

The resistance of R3 is given by:

$$R_3 = \frac{5}{1} = 5 \, \Omega$$

The total resistance of the parallel circuit (R2 and R3) is given by:

$$\frac{1}{R_{\text{parallel}}} = \frac{1}{5} + \frac{1}{5}$$

$$R_{\text{parallel}} = 2.5 \, \Omega$$

Since the total resistance of the circuit was determined to be 5 \Omega, the resistance of R1 is:

$$R_1 = 5 - 2.5 = 2.5 \, \Omega$$