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## Q. 5.31

a. For the series circuit of Fig. 5.67 find the voltages $V_a, V_b, and V_c$.
b. Determine the voltage $V_{ad}$.

## Verified Solution

a. Clearly,

$V_a = 0 V$

The current:

$I=\frac{E_1-E_2}{R_1+R_2+R_3}=\frac{240V-60V}{10\Omega +40\Omega +50\Omega } =\frac{180V}{100\Omega }=1.8A$

so that                       $V_2=IR_2=(1.8A)(40\Omega )=72V\\[0.5cm] V_3=IR_3=(1.8A)(50\Omega )=90V$

and                            $V_b=-V_2-V_3=-72V-90V=162V\\[0.5cm] V_c=-V_2=-(72V)=-72V$

b. Applying Kirchhoff’s voltage law:

$V_{ad}+V_1-E_1=0$

so that                 $V_{ad}=E_1-V_1$

with                     $V_1=I(R_1)=(1.8A)(10\Omega )=18V$

and finally         $V_{ad}=E_1-V_1=240V-18V=222V$