In the circuit below, I have to calculate \$V_u\$.


simulate this circuit – Schematic created using CircuitLab

By applying the voltage divider between the series resistances below, I get $$V_u=V_e\frac{R}{R+R}-V_e\frac{R}{R+R}=0$$ However if I apply the voltage divider to the two elements above, I get $$V_u=V_e\frac{R}{R+\frac{1}{j\omega C}}-V_e\frac{\frac{1}{j\omega C}}{R+\frac{1}{j\omega C}}$$

The solution gives directly:

$$V_u=V_e\frac{R}{R+\frac{1}{j\omega C}}-V_e\frac{R}{R+R}$$

Why? How did it apply the formula?

  • \$\begingroup\$ This is basically the same question as this one, except that this has an extra resistor instead of a capacitor. My answer explains the process for finding the differential voltage (\$V_u\$ in your case). \$\endgroup\$
    – Null
    May 26 '15 at 16:55

The potential difference \$V_u\$ is the difference between the voltage drops across the two resistors in the two legs. Hench they have applied 2 voltage dividers to find the potentials across each of the resistors.

$$V_{R2}=V_e\times\frac{R}{R+\frac{1}{jwC}} = \text{potential across the resistor in the second leg} $$

$$V_{R1} = V_e \times \frac{R}{R+R}=\text{Potential across the resistor in the first leg}$$

\$V_u\$ is simply the difference between the potentials above.

In the method you have used you are mistaking what \$V_u\$ stands for and thats where you are going wrong.


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