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## Q. 5.26

Determine $V_{ab}, V_{cb}, and V_c$ for the network in Fig. 5.59. ## Verified Solution

There are two ways to approach this problem. The first is to sketch the diagram in Fig. 5.60 and note that there is a 54 V drop across the series resistors $R_1 and R_2$. The current can then be determined using Ohm’s law and the voltage levels as follows:

$I=\frac{54V}{45\Omega }=1.2A$

$V_{ab}=IR_2=(1.2A)(25\Omega )=30V$

$V_{cb}=-IR_1=-(1.2A)(20\Omega )=-24V$

$V_c=E_1=-19V$

The other approach is to redraw the network as shown in Fig. 5.61 to clearly establish the aiding effect of $E_1 and E_2$ and then solve the resulting series circuit:

$I=\frac{E_1+E_2}{R_T}=\frac{19V+35V}{45\Omega } =\frac{54V}{45\Omega } =1.2A$

and        $V_{ab}=30V\qquad V_{cb}=-24V\qquad V_c=-19V$  