Electrical Engineering BMT & UFC › Circuit Analysis 2

Simplifying Circuits

A real circuit generally consists of series and parallel connections of many components. Following the rules just worked out, these can be combined further and further step by step. Combining all components of a circuit is used, for example, to determine the current load on a voltage source.

Load on a source

When you select a battery to supply a circuit, the manufacturer specifies in the data sheet a maximum current that can be drawn from the battery. Evidently, a car battery with a volume of several litres can deliver more current than a small button cell in a wristwatch. The current consumption of the circuit – i.e. the current that flows from the battery into the circuit – must therefore match the battery.

It is mathematically not easy to determine the current consumption of a complex circuit. If the circuit is reduced to, at best, a single equivalent resistor, the calculation is easy.

This is shown using the example of the following circuit, which is to operate several loads, all modelled as ohmic resistors. The following applies

Example circuit with R1 to R5 on a LiPo battery

\[ \begin{gathered} R_1 = 4.4\,\Omega,\ R_2 = 6\,\Omega,\ R_3 = R_4 = R_5 = 4\,\Omega \\[4pt] \text{At the LiPo battery: } U_0 = 3.7\,\mathrm{V} \end{gathered} \]

To calculate the load on the voltage source with the current I₀, the circuit is simplified step by step without changing the voltage and current at the source.

In the first step, R₄ and R₅, which are connected in parallel, are combined into one resistor. This results in the following simplified circuit:

\[ R_{45} = R_4 \,||\, R_5 = \frac{4\,\Omega \cdot 4\,\Omega}{4\,\Omega + 4\,\Omega} = 2\,\Omega \]

In the second step, the series connection of R₃ and R₄₅ is combined:

\[ R_{345} = R_3 + R_{45} = 4\,\Omega + 2\,\Omega = 6\,\Omega \]

Now R₂ and R₃₄₅ are in parallel. In a further simplification step, the following applies:

\[ R_{2345} = R_2 \,||\, R_{345} = \frac{6\,\Omega \cdot 6\,\Omega}{6\,\Omega + 6\,\Omega} = 3\,\Omega \]

Finally, the series connection of R₁ and R₂₃₄₅ is combined into a total resistance.

Total resistance RGes at the source — Ges = total

\[ R_{\mathrm{Ges}} = R_1 + R_{2345} = 4.4\,\Omega + 3\,\Omega = 7.4\,\Omega \]

The current loading the source can now easily be calculated as

\[ I_0 = \frac{U_0}{R_{\mathrm{Ges}}} = \frac{3.7\,\mathrm{V}}{7.4\,\Omega} = 0.5\,\mathrm{A} \]

This current flows from the battery into the circuit. With this value, the battery can now be selected.

Calculating partial voltages and partial currents

The different simplification steps can be used backwards to calculate partial currents and partial voltages in a network more easily. For this purpose, the partial voltage U₂ across R₂ and the partial current I₃ flowing through R₃ are calculated in the example above.

Example circuit with the required quantities U2 and I3

The voltage U₂ equals the potential difference between the nodes above and below the resistor R₂. So U₂ is present across the equivalent resistor R₂₃₄₅ in the second-to-last simplification step of the circuit.

The current I₀ = 0.5 A flows through R₂₃₄₅. According to Ohm's law, the following voltage drops across this resistor:

Voltage divider consisting of R1 and R2345

\[ U_2 = R_{2345} \cdot I_0 = 3\,\Omega \cdot 0.5\,\mathrm{A} = 1.5\,\mathrm{V} \]

The current I₃ can be calculated relatively easily using a current divider in the third simplification step, because this current flows through the equivalent resistor R₃₄₅:

Current divider consisting of R2 and R345

\[ \text{Current divider: } I_3 = I_0 \cdot \frac{R_2}{R_2 + R_{345}} = 0.5\,\mathrm{A} \cdot \frac{6\,\Omega}{6\,\Omega + 6\,\Omega} = 0.5\,\mathrm{A} \cdot \frac{1}{2} = 0.25\,\mathrm{A} \]

Simulation