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# Combined Series-Parallel Circuits

## Representing most real world circuits, these circuits are connected in series as well as in parallel.

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Combined Series-Parallel Circuits

Electrical circuits can become immensely complicated. This circuit is a polynomial plotter, which allows users to plot polynomials and evaluate functions at various x\begin{align*}x\end{align*} values.

### Combined Series-Parallel Circuits

Most circuits are not just a series or parallel circuit; most have resistors in parallel and in series. These circuits are called combination circuits. When solving problems with such circuits, use this series of steps.

1. For resistors connected in parallel, calculate the single equivalent resistance that can replace them.
2. For resistors in series, calculate the single equivalent resistance that can replace them.
3. By repeating steps 1 and 2, you can continually reduce the circuit until only a single equivalent resistor remains. Then you can determine the total circuit current. The voltage drops and currents though individual resistors can then be calculated.

Example Problem: In the combination circuit sketched below, find the equivalent resistance for the circuit, find the total current through the circuit, and find the current through each individual resistor.

Solution: We start by simplifying the parallel resistors R2\begin{align*}R_2\end{align*} and R3\begin{align*}R_3\end{align*}.

1R23R23=1180 Ω+1220 Ω=199 Ω=99 Ω

We then simplify R1\begin{align*}R_1\end{align*} and R23\begin{align*}R_{23}\end{align*} which are series resistors.

RT=R1+R23=110 Ω+99 Ω=209 Ω\begin{align*}R_T=R_1+R_{23}=110 \ \Omega + 99 \ \Omega = 209 \ \Omega\end{align*}

We can then find the total current, IT=VTRT=24 V209 Ω=0.11 A\begin{align*}I_T=\frac{V_T}{R_T}=\frac{24 \ V}{209 \ \Omega}=0.11 \ A\end{align*}

All the current must pass through R1\begin{align*}R_1\end{align*}, so I1=0.11 A\begin{align*}I_1=0.11 \ A\end{align*}.

The voltage drop through R1\begin{align*}R_1\end{align*} is (110 Ω)(0.11 A)=12.6 volts\begin{align*}(110 \ \Omega) (0.11 \ A)=12.6 \text{ volts}\end{align*}.

Therefore, the voltage drop through R2\begin{align*}R_2\end{align*} and R3\begin{align*}R_3\end{align*} is 11.4 volts.

I2=V2R2=11.4 V180 Ω=0.063 A\begin{align*}I_2=\frac{V_2}{R_2}=\frac{11.4 \ V}{180 \ \Omega}=0.063 \ A\end{align*} and I3=V3R3=11.4 V220 Ω=0.052 A\begin{align*}I_3=\frac{V_3}{R_3}=\frac{11.4 \ V}{220 \ \Omega}=0.052 \ A\end{align*}

#### Summary

• Combined circuit problems should be solved in steps.

#### Practice

Questions

Video teaching the process of simplifying a circuit that contains both series and parallel parts.

1. In a circuit that contains both series and parallel parts, which parts of the circuit are simplified first?
2. In the circuit drawn below, which resistors should be simplified first?

#### Review

Questions

1. Two 60.0Ω resistors are connected in parallel and this parallel arrangement is then connected in series with a 30.0Ω resistor. The combination is placed across a 120. V potential difference.
1. Draw a diagram of the circuit.
2. What is the equivalent resistance of the parallel portion of the circuit?
3. What is the equivalent resistance for the entire circuit?
4. What is the total current in the circuit?
5. What is the voltage drop across the 30.0Ω resistor?
6. What is the voltage drop across the parallel portion of the circuit?
7. What is the current through each resistor?
2. Three 15.0 Ohm resistors are connected in parallel and the combination is then connected in series with a 10.0 Ohm resistor. The entire combination is then placed across a 45.0 V potential difference. Find the equivalent resistance for the entire circuit.