10.2 Resistors in series and parallel  (Page 7/11)

Looking at [link] (c), this leaves $24\text{V}-14\text{V}=10\text{V}$ to be dropped across the parallel combination of $R_2$ and $R_{34}.$ The current through $R_2$ can be found using Ohm’s law:

The resistors $R_3$ and $R_4$ are in series so the currents $I_3$ and $I_4$ are equal to

Using Ohm’s law, we can find the potential drop across the last two resistors. The potential drops are $V_3=I_3{R}_3=6\text{V}$ and $V_4=I_4{R}_4=4\text{V}.$ The final analysis is to look at the power supplied by the voltage source and the power dissipated by the resistors. The power dissipated by the resistors is

The total energy is constant in any process. Therefore, the power supplied by the voltage source is $P_s=IV=(2\text{A})(24\text{V})=48\text{W}.$ Analyzing the power supplied to the circuit and the power dissipated by the resistors is a good check for the validity of the analysis; they should be equal.

Combining series and parallel circuits

Strategy

(b) The current through $R_1$ can be found using Ohm’s law and the voltage applied. The current through $R_1$ is equal to the current from the battery. The potential drop $V_1$ across the resistor $R_1$ (which represents the resistance in the connecting wires) can be found using Ohm’s law.

(c) The current through $R_2$ can be found using Ohm’s law $I_2=\frac{V_2}{R_2}.$ The voltage across $R_2$ can be found using $V_2=V-V_1.$

(d) Using Ohm’s law $\left(V_2=I_2{R}_2\right)$ , the power dissipated by the resistor can also be found using $P_2=I_2^2{R}_2=\frac{V_2^2}{R_2}$ .

Solution

1. To find the equivalent resistance of the circuit, notice that the parallel connection of $R_2$ and $R_3$ is in series with $R_1$ , so the equivalent resistance is
   
   $R_{\text{eq}}=R_1+{\left(\frac{1}{R_2}+\frac{1}{R_3}\right)}^{\mathrm{-1}}=1.00\text{Ω}+{\left(\frac{1}{6.00\text{Ω}}+\frac{1}{13.00\text{Ω}}\right)}^{\mathrm{-1}}=5.10\text{Ω}.$
   
   The total resistance of this combination is intermediate between the pure series and pure parallel values ( $20.0\text{Ω}$ and $0.804\text{Ω}$ , respectively).
2. The current through $R_1$ is equal to the current supplied by the battery:
   
   $I_1=I=\frac{V}{R_{\text{eq}}}=\frac{12.0\text{V}}{5.10\text{Ω}}=2.35\text{A}.$
   
   The voltage across $R_1$ is
   
   $V_1=I_1{R}_1=(2.35\text{A})(1\text{Ω})=2.35\text{V}.$
   
   The voltage applied to $R_2$ and $R_3$ is less than the voltage supplied by the battery by an amount $V_1.$ When wire resistance is large, it can significantly affect the operation of the devices represented by $R_2$ and $R_3$ .
3. To find the current through $R_2$ , we must first find the voltage applied to it. The voltage across the two resistors in parallel is the same:
   
   $V_2=V_3=V-V_1=12.0\text{V}-2.35\text{V}=9.65\text{V}.$
   
   Now we can find the current $I_2$ through resistance $R_2$ using Ohm’s law:
   
   $I_2=\frac{V_2}{R_2}=\frac{9.65\text{V}}{6.00\text{Ω}}=1.61\text{A}.$
   
   The current is less than the 2.00 A that flowed through $R_2$ when it was connected in parallel to the battery in the previous parallel circuit example.
4. The power dissipated by $R_2$ is given by
   
   $P_2=I_2^2{R}_2={(1.61\text{A})}^2(6.00\text{Ω})=15.5\text{W}.$