Combining Series and Parallel: The Master Circuit Analysis

As circuits become more complex, you will find that resistors (or other components) come in a mix that combines series and parallel connections together. Although you often won't need to simplify such complex circuits to arrive at a single total resistance in fieldwork, it is very common to deal with circuits that combine sections of series and sections of parallel. It is important here to track the branching points of the circuit to monitor the split current paths; the key is to determine where the current paths branch. When two components are connected and there is only one path for the current, the two components are in series. However, if the path splits into two or more paths, the components are connected in parallel.

In the first circuit described in the text, the positive terminal of the battery is connected only to the resistor R_1 and is not connected anywhere else; thus, there is only one path for the current at that point. On the other side of R_1, the circuit splits into two paths for the current, one passing through R_2 and the other through R_3. Since we now have two paths for the current, we are dealing with a parallel circuit consisting of R_2 and R_3. These two resistors are connected in parallel with each other, but together (as a single group) they are connected in series with R_1.

The second circuit illustrated in the text is electrically identical to the first; the circuit is not classified as a series or parallel circuit based on the way the schematic diagram is drawn, but rather based on how the components are connected in reality. In the second circuit, $R_3$ is still connected in parallel with $R_2$. In both circuits, the current path splits at the junction of $R_1$ with $R_2$ and $R_3$, and then rejoins at the connection point of the opposite terminals of $R_2$ and $R_3$ together.

The left side of this circuit represents a series circuit (one path for current), while the right side represents a parallel circuit (multiple paths for current). The following diagram illustrates how the rules of series and parallel circuits merge together within a compound circuit (series-parallel).

1. Current Analysis

First, let's look at the current: just like in a series circuit, a total current of 3 amperes flows around the entire circuit. However, at the branching point of the path, 2 amperes out of the 3 amperes go through the 15-ohm resistor, while the remaining 1 ampere passes through the 30-ohm resistor (where the larger current flows through the smaller resistance). The currents meet and combine again at the junction of these two resistors to return to their original total value of 3 amperes, which perfectly satisfies Kirchhoff's Current Law (KCL).

2. Voltage Analysis

After that, let's take a look at the efforts:

The two resistors (15 ohms and 30 ohms) are connected in parallel with each other, resulting in a total equivalent resistance of 10 ohms. This equivalent resistance is considered to be connected in series with the first 10-ohm resistor. If we look at the circuit from a "general and simplified perspective," it will behave as if there are two 10-ohm resistors connected in series. From here, we see Kirchhoff's Voltage Law (KVL) in actual operation, where the voltage across the parallel combination (30 volts) is added to the voltage across the 10-ohm resistor (30 volts), resulting in a total equal to the source voltage of 60 volts.

The battery supplies the circuit with 60 volts; half of this voltage is lost across the first 10-ohm resistor ($10\ \Omega \times 3\text{ A} = 30\text{ V}$). The other 30 volts are lost across the combination of the two parallel resistors (15 ohms and 30 ohms). You can verify this for each resistor individually:

• Through the first parallel resistor: $15\ \Omega \times 2\text{ A} = 30\text{ V}$
• Across the second parallel resistor: $30\ \Omega \times 1\text{ A} = 30\text{ V}$

You can also prove this by treating the two resistors together as a single block (30 ohms in parallel with 15 ohms = 10 ohms, and when multiplying 10 ohms by the total current of 3 amperes, we get 30 volts). The two resistors connected in parallel behave like a single resistor with a value of 10 ohms, and applying Ohm's Law still yields the same voltage exactly.

💡 Technical Deep Dive (For Technicians and Engineers)

1. Do not add parallel voltages: Be careful, and do not forget the rules of the parallel circuit; do not add the 30 volts generated across the 15-ohm resistor with the 30 volts generated across the 30-ohm resistor to say that the result is 60 volts! It is the same 30 volts that appear across both resistors together as a single group. 30 volts are lost across the 10-ohm resistor, and then another 30 volts are lost across the parallel group, which equals the 60 volts supplied by the battery and satisfies Kirchhoff's voltage law.
2. Do not apply the full source voltage to a single component: Do not fall into the common mistake of applying the full battery voltage (60 volts) directly across the first 10-ohm resistor; it is connected in series with the rest of the circuit, and therefore the voltages will be different (series circuit rule: voltages are distributed and differ). Do not apply the voltage present in one part of a series circuit to another part of it.

When combining the 15-ohm and 30-ohm resistors, we get an equivalent resistance of 10 ohms. Now, the source voltage of 60 volts is divided equally between the two equal resistors (the first 10 ohm and the 10 ohm equivalent in parallel) connected in series. At this stage, you can correctly calculate the current; whether you use the 30 volts across one of the 10-ohm resistors or the full 60 volts across the total resistance (20 ohms), you will get a current of 3 amperes in both cases.

🛠️ Solve the practical exercise mentioned in the text:

Based on the data and questions provided in the circuit attached to the exercise, here are the precise scientific answers:

• What is the equivalent resistance of R2 and R3? It equals 10 ohms (the product of them divided by the sum of them: $(15 \times 30) / (15 + 30) = 450 / 45 = 10\ \Omega$).
• What is the total resistance of the circuit? It is equal to 20 ohms; because the entire circuit behaves as if it were two resistors of 10 ohms connected in series ($10\ \Omega + 10\ \Omega = 20\ \Omega$).
• What is the voltage across R1? It is equal to 10 volts (or half of the total available voltage for that stage based on the series ratios).
• What is the voltage across R2 equal to 10 volts (because the parallel group shares the same voltage)?
  
• What is the current flowing through R2?0.5 A.
• What is the current flowing through R3?0.5 A.
• What is the current flowing through R1? It equals 1 ampere total (which is the sum of the two branching currents $0.5 + 0.5 = 1\text{ A}$ that passes through the resistor connected in series).