We were given the task of finding the voltage drop across R1 of the circuit below with the following alternatives:

  1. 5.8V

  2. 4.8V

  3. 4.0V

  4. 10.0V

Some Circuit

All given information is contained within the image, i.e. no frequency is specified.

My intuition tells me that the only way to solve this is to consider the capacitors as resistors with a resistance equal the given reactance and solve it as a DC circuit where the voltage drop over R1 becomes 3.2V, i.e. none of the above.

I have a hunch that the alternatives are wrong. However, if anyone have any ideas on how to solve this, let me know.

The correct answer is apparently 5.8V.

  • \$\begingroup\$ I would say you are correct with your 3.2V answer. It is impossible that the lowest resistor value in the circuit drops more than half of the supplied voltage... \$\endgroup\$ – Douwe66 Jan 7 '19 at 9:36
  • 2
    \$\begingroup\$ Not only do you not have frequencies, you don't have capacitance values. So, you pretty much have to do it using the given reactance, and ignore the frequency. But, you have to keep in mind that you are dealing with the reactance of capacitors. How does that affect the reactances (summing and parallel circuits?) \$\endgroup\$ – JRE Jan 7 '19 at 9:56
  • \$\begingroup\$ The reactances are given so you do not need frequency. You cannot treat reactances as resistances. Resistances and Reactances must be combined into impedances. For example: Z1 = R1 - jXC1 \$\endgroup\$ – StainlessSteelRat Jan 7 '19 at 17:01

First of all you don't care about the frequency of the source, since you have the reactance of the caps:

\$X_C=-j\frac{1}{ωC} = -j \cdot\texttt{fixed}\$

\$R_2\$ and \$C_2\$ are in parallel, so:

\$R_2//X_{C_2} = \frac{R_2\cdot X_{C_2}}{R_2+ X_{C_2}}=\frac{333\cdot (-j250)}{333-j250}=\frac{83250\angle{-90}}{416.4\angle{-35.9}}=200\angle{-53.1}\$

Then, \$R_1, C_1\$ and \$R_2//X_{C_2}\$ are in series. Calculating the total Z can give you the current exiting the source and finally \$V_{R_1}=R1\cdot I\$.

Keep in mind that books make mistakes too...


As the impedance of each component is given, the frequency of the supply does not matter.

The Voltage Drop across \$R1\$ will be found by first calculating total impedance of \$R2\$, \$C1\$ and \$C2\$ As \$R2//XC2\$ therefore, \$R2//XC2=\frac{R2⋅XC2}{R2+XC2}=\frac{333⋅(−j250)}{333−j250}=200\angle−53.1\$

then \$XC1+(R2//XC2)=240\angle-90+200\angle-53.1=120.1-j399.94\$

Suppose the sum is denoted by \$Zo\$. The voltage across R1 will be calculated by the voltage divider rule which is

$$ V_{R1}=V_{s}*\frac{R1}{R1+Zo} $$


You can solve this as a voltage divider, but you have to deal with the impedances as complex numbers. The given impedance for the capacitors should be taken as -j|Xc| where Xc is what has been given.

So the magnitude of the voltage output will simply be:

|Vout| = \$10 \cdot |\frac{Z1}{R1-jXc1+Z1}|\$ where Z1 = \$\frac{R2\cdot (-jXc2)}{R2 -jXc2}\$

The rest of it is just straightforward arithmetic with complex numbers.

Remember to multiply numerator and denominator by the complex conjugate of the denominator to simplify complex ratios (assuming you are working in rectangular coordinates), and how to find the magnitude of a complex number.

Hint: The supposed answer is not correct, however the correct answer is in the list of possible answers.


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