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## 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 $V_{ab}$. ## 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                        $V_b=-32V$

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