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Chapter 5

Q. 5.30

For the series network of Fig. 5.66, determine
a. The voltage V_a.
b. The voltages V_b  and  V_c.
c. The voltage V_{ab}.

Step-by-Step

Verified Solution

a. Let us first determine the current:

I=\frac{E}{R_1+R_2+R_3}=\frac{72V}{6\Omega +8\Omega +4\Omega } =\frac{72V}{18\Omega }=4A

The voltage V_a is then

V_a=V_1=IR_1=(4A)(6\Omega )=24V

b. The voltage V_b is equal to

V_b=-V_2

with                            V_2=IR_2=(4A)(8\Omega )=32V

so that                        V_b=-32V

V_c=-V_2-V_3=-32V-(I)(R_3)\\[0.5cm]=-32V-(4A)(4\Omega )\\[0.5cm] =-32V-16V\\[0.5cm] =-48V