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Chapter 5

Q. 5.26

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

Step-by-Step

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