Circuit Analysis Using Series/Parallel Equivalents
Find the current, voltage, and power for each element of the circuit shown in Figure 2.5(a).

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First, we combine resistances in series and parallel. For example, in the original circuit, [latex]R_2[/latex] and [latex]R_3[/latex] are in parallel. Replacing [latex]R_2[/latex] and [latex]R_3[/latex] by their parallel equivalent, we obtain the circuit shown in Figure 2.5(b). Next, we see that [latex]R_1[/latex] and [latex]R_{eq1}[/latex] are in series. Replacing these resistances by their sum, we obtain the circuit shown in Figure 2.5(c).
After we have reduced a network to an equivalent resistance connected across the source, we solve the simplified network. Then, we transfer results back through the chain of equivalent circuits. We illustrate this process in Figure 2.6. (Figure 2.6 is identical to Figure 2.5, except for the currents and voltages shown in Figure 2.6.

Usually, in solving a network by this technique, we first draw the chain of equivalent networks and then write results on the same drawings. However, this might be confusing in our first example.)
First, we solve the simplified network shown in Figure 2.6(c). Because [latex]R_{eq}[/latex] is parallel with the 90-V voltage source, the voltage across [latex]R_{eq}[/latex] must be 90 V, with its positive polarity at the top end. Thus, the current owing through [latex]R_{eq}[/latex] is given by

[latex]i_1=\frac{v_s}{R_{eq}} =\frac{90\mathrm{~V}}{30~\Omega} =3\mathrm{~A}[/latex]

We know that this current flows downward (from plus to minus) through [latex]R_{eq}[/latex]. Since [latex]v_s[/latex] and [latex]R_{eq}[/latex] are in series in Figure 2.6(c), the current must also flow upward through [latex]v_s[/latex]. Thus,[latex]i_1=3\mathrm{~A}[/latex] flows clockwise around the circuit, as shown in Figure 2.6(c).
Because [latex]R_{eq}[/latex] is the equivalent resistance seen by the source in all three parts of Figure 2.6, the current through [latex]v_s[/latex] must be [latex]i_1=3\mathrm{~A}[/latex], flowing upward in all three equivalent circuits. In Figure 2.6(b), we see that [latex]i_1[/latex] flows clockwise through [latex]v_s,R_1,[/latex] and [latex]R_{eq1}[/latex]. The voltage across [latex]R_{eq1}[/latex] is given by

[latex]v_2=R_{qe1}i_1=20~\Omega imes 3\mathrm{~A}=60\mathrm{~V}[/latex]

Because [latex]R_{eq1}[/latex] is the equivalent resistance for the parallel combination of [latex]R_{2}[/latex] and [latex]R_{3}[/latex], the voltage [latex]v_{2}[/latex] also appears across [latex]R_{2}[/latex] and [latex]R_{3}[/latex] in the original network.
At this point, we have found that the current through [latex]v_s[/latex] and [latex]R_{1}[/latex] is [latex]i_1=3\mathrm{~A}[/latex]. Furthermore, the voltage across [latex]R_{2}[/latex] and [latex]R_{3}[/latex] is 60 V. This information is shown in Figure 2.6(a). Now, we can compute the remaining values desired:

[latex]i_2=\frac{v_2}{R_2}=\frac{60\mathrm{~V}}{30~\Omega} =2\mathrm{~A}[/latex]

[latex]i_3=\frac{v_2}{R_3}=\frac{60\mathrm{~V}}{60~\Omega} =1\mathrm{~A}[/latex]

(As a check, we can use KCL to verify that [latex]i_1=i_2+i_3.[/latex])
Next, we can use Ohm s law to compute the value of [latex]v_1[/latex]:

[latex]v_1=R_1i_1=10~\Omega imes 3\mathrm{~A}=30\mathrm{~V}[/latex]

(As a check, we use KVL to verify that [latex]v_s=v_1+v_2.[/latex])
Now, we compute the power for each element. For the voltage source, we have

[latex]p_s=-v_si_1[/latex]

We have included the minus sign because the references for [latex]v_s[/latex] and [latex]i_1[/latex] are opposite to the passive configuration. Substituting values, we have

[latex]p_s=-\left(90\mathrm{~V} ight) imes 3\mathrm{~A}=-270\mathrm{~W}[/latex]

Because the power for the source is negative, we know that the source is supplying energy to the other elements in the circuit.
The powers for the resistances are

[latex]p_1=R_1i_1^2=10~\Omega imes \left(3\mathrm{~A} ight)^2=90\mathrm{~W}[/latex]

[latex]p_2=\frac{v_2^2}{R_2}=\frac{\left(60\mathrm{~V} ight)^2 }{30~\Omega}=120\mathrm{~W} [/latex]

[latex]p_3=\frac{v_2^2}{R_3}=\frac{\left(60\mathrm{~V} ight)^2 }{60~\Omega}=60\mathrm{~W} [/latex]

(As a check, we verify that [latex]p_s+p_1+p_2+p_3=0[/latex], showing that power is conserved.)

Question 2.2: Circuit Analysis Using Series/Parallel Equivalents Find the ...

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