3.5 Series and parallel circuits  (Page 2/3)

One interesting simple circuit ( [link] ) has two resistors connected side-by-side, what we will term a parallel connection, rather than in series. Here, applying KVL reveals that all the voltages are identical: $v_1=v$ and $v_2=v$ . This result typifies parallel connections. To write the KCLequation, note that the top node consists of the entire upper interconnection section. The KCL equation is $i_{\mathrm{in}}-i_1-i_2=0$ . Using the v-i relations, we find that $$i_{\mathrm{out}}=\frac{R_1}{R_1+R_2}{i}_{\mathrm{in}}$$

This circuit highlights some important properties of parallel circuits. You can easily show that the parallel combination of $R_1$ and $R_2$ has the v-i relation of a resistor having resistance $(\frac{1}{R_1}+\frac{1}{R_2})^{-1}=\frac{R_1{R}_2}{R_1+R_2}$ . A shorthand notation for this quantity is $\parallel (R_1, R_2)$ . As the reciprocal of resistance is conductance , we can say that for a parallel combination of resistors, the equivalent conductance is the sum of the conductances .

Similar to voltage divider for series resistances, we have current divider for parallel resistances. The current through a resistor in parallel with another is the ratio of theconductance of the first to the sum of the conductances. Thus, for the depicted circuit, $i_2=\frac{G_2}{G_1+G_2}i$ . Expressed in terms of resistances, current divider takes the form of the resistance of the other resistor divided by the sum of resistances: $i_2=\frac{R_1}{R_1+R_2}i$ .

Suppose we want to pass the output signal into a voltage measurement device, such as an oscilloscope or a voltmeter. Insystem-theory terms, we want to pass our circuit's output to a sink. For most applications, we can represent these measurementdevices as a resistor, with the current passing through it driving the measurement device through some type of display.In circuits, a sink is called a load ; thus, we describe a system-theoretic sink as a load resistance $R_L$ . Thus, we have a complete system built from a cascade of threesystems: a source, a signal processing system (simple as it is), and a sink.

We must analyze afresh how this revised circuit, shown in [link] , works. Rather than defining eight variables and solving for thecurrent in the load resistor, let's take a hint from other analysis( series rules , parallel rules ). Resistors $R_2$ and $R_L$ are in a parallel configuration: The voltages across each resistor are the same while the currents arenot. Because the voltages are the same, we can find the current through each from their v-i relations: $i_2=\frac{v_{\mathrm{out}}}{R_2}$ and $i_L=\frac{v_{\mathrm{out}}}{R_L}$ . Considering the node where all three resistors join, KCL saysthat the sum of the three currents must equal zero. Said another way, the current entering the node through $R_1$ must equal the sum of the other two currents leaving the node. Therefore, $i_1=i_2+i_L$ , which means that $i_1=v_{\mathrm{out}}(\frac{1}{R_2}+\frac{1}{R_L})$ .