10.2 Resistors in series and parallel  (Page 9/11)

Combining series and parallel circuits

Strategy

Solution

1. Draw a clear circuit diagram ( [link] ).
   

   

2. The unknown is the voltage of the battery. In order to find the voltage supplied by the battery, the equivalent resistance must be found.
3. In this circuit, we already know that the resistors $R_1$ and $R_2$ are in series and the resistors $R_3$ and $R_4$ are in parallel. The equivalent resistance of the parallel configuration of the resistors $R_3$ and $R_4$ is in series with the series configuration of resistors $R_1$ and $R_2$ .
4. The voltage supplied by the battery can be found by multiplying the current from the battery and the equivalent resistance of the circuit. The current from the battery is equal to the current through $R_1$ and is equal to 2.00 A. We need to find the equivalent resistance by reducing the circuit. To reduce the circuit, first consider the two resistors in parallel. The equivalent resistance is $R_{34}={\left(\frac{1}{10.00\text{Ω}}+\frac{1}{10.00\text{Ω}}\right)}^{\mathrm{-1}}=5.00\text{Ω}.$ This parallel combination is in series with the other two resistors, so the equivalent resistance of the circuit is $R_{\text{eq}}=R_1+R_2+R_{34}=25.00\text{Ω}.$ The voltage supplied by the battery is therefore $V=I{R}_{\text{eq}}=2.00\text{A}\left(25.00\text{Ω}\right)=50.00\text{V}.$
5. One way to check the consistency of your results is to calculate the power supplied by the battery and the power dissipated by the resistors. The power supplied by the battery is $P_{\text{batt}}=IV=100.00\text{W}.$
   Since they are in series, the current through $R_2$ equals the current through $R_1.$ Since $R_3=R_4$ , the current through each will be 1.00 Amps. The power dissipated by the resistors is equal to the sum of the power dissipated by each resistor:
   
   $P=I_1^2{R}_1+I_2^2{R}_2+I_3^2{R}_3+I_4^2{R}_4=40.00\text{W}+40.00\text{W}+10.00\text{W}+10.00\text{W}=100.00\text{W}.$
   
   Since the power dissipated by the resistors equals the power supplied by the battery, our solution seems consistent.

Significance

Summary

• The equivalent resistance of an electrical circuit with resistors wired in a series is the sum of the individual resistances: $R_{\text{s}}=R_1+R_2+R_3+\text{⋯}={\displaystyle \sum _{i=1}^N{R}_i}$ .
• Each resistor in a series circuit has the same amount of current flowing through it.
• The potential drop, or power dissipation, across each individual resistor in a series is different, and their combined total is the power source input.
• The equivalent resistance of an electrical circuit with resistors wired in parallel is less than the lowest resistance of any of the components and can be determined using the formula
  
  $R_{\text{eq}}={\left(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\text{⋯}\right)}^{\mathrm{-1}}={\left({\displaystyle \sum _{i=1}^N\frac{1}{R_i}}\right)}^{\mathrm{-1}}.$
• Each resistor in a parallel circuit has the same full voltage of the source applied to it.
• The current flowing through each resistor in a parallel circuit is different, depending on the resistance.
• If a more complex connection of resistors is a combination of series and parallel, it can be reduced to a single equivalent resistance by identifying its various parts as series or parallel, reducing each to its equivalent, and continuing until a single resistance is eventually reached.