10.2 Resistors in series and parallel  (Page 2/11)

This equation is often referred to as Kirchhoff’s loop law, which we will look at in more detail later in this chapter. For [link] , the sum of the potential drop of each resistor and the voltage supplied by the voltage source should equal zero:

Since the current through each component is the same, the equality can be simplified to an equivalent resistance, which is just the sum of the resistances of the individual resistors.

Any number of resistors can be connected in series. If N resistors are connected in series, the equivalent resistance is

One result of components connected in a series circuit is that if something happens to one component, it affects all the other components. For example, if several lamps are connected in series and one bulb burns out, all the other lamps go dark.

Equivalent resistance, current, and power in a series circuit

Strategy

Solution

1. The equivalent resistance is the algebraic sum of the resistances:
   
   $R_{\text{eq}}=R_1+R_2+R_3+R_4+R_5=20\text{Ω}+20\text{Ω}+20\text{Ω}+20\text{Ω}+10\text{Ω}=90\text{Ω}.$
2. The current through the circuit is the same for each resistor in a series circuit and is equal to the applied voltage divided by the equivalent resistance:
   
   $I=\frac{V}{R_{\text{eq}}}=\frac{9\text{V}}{90\text{Ω}}=0.1\text{A.}$
3. The potential drop across each resistor can be found using Ohm’s law:
   
   $\begin{array}{c}{V}_1=V_2=V_3=V_4=\left(0.1\text{A}\right)20\text{Ω}=2\text{V},\hfill \\ V_5=\left(0.1\text{A}\right)10\text{Ω}=1\text{V},\hfill \\ V_1+V_2+V_3+V_4+V_5=9\text{V}.\hfill \end{array}$
   
   Note that the sum of the potential drops across each resistor is equal to the voltage supplied by the battery.
4. The power dissipated by a resistor is equal to $P=I^{2}R$ , and the power supplied by the battery is equal to $P=I\epsilon$ :
   
   $\begin{array}{}\\ P_1=P_2=P_3=P_4={\left(0.1\text{A}\right)}^2\left(20\text{Ω}\right)=0.2\text{W},\hfill \\ P_5={\left(0.1\text{A}\right)}^2\left(10\text{Ω}\right)=0.1\text{W},\hfill \\ P_{\text{dissipated}}=0.2\text{W}+0.2\text{W}+0.2\text{W}+0.2\text{W}+0.1\text{W}=0.9\text{W},\hfill \\ P_{\text{source}}=I\epsilon =\left(0.1\text{A}\right)\left(9\text{V}\right)=0.9\text{W}.\hfill \end{array}$

Significance