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I'm a new student to electrical and electronics engineering. I've been given the following calculations regarding an A.C. circuit with resistors and a capacitor in series, and I'm not sure if I've answered correctly. Any help is much appreciated!

Question:

For the circuit shown below calculate the: i) Reactance of capacitor C1. ii) Impedance of the circuit using a phasor diagram. iii) Power dissipation for resistor R2.

Circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

My answers:

i) \$X_C = \frac {1} {(2\pi \cdot 50Hz \cdot 1.5\cdot10^{-5}F)} = 212.2Ω\$

ii) \$Z = \sqrt {(150\Omega)^2 + (212.2\Omega)^2} = 259.9\Omega\$

iii) \$I = \frac {1V} {(150\Omega + 212.2\Omega)} \approx 0.00276A = 2.76mA\$

\$P = I^2 R = (2.76mA)^2 \cdot 50\Omega = 0.14W\$

What do you think? Thanks.

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  • \$\begingroup\$ I = VS / Z, which effects your P calc also. Z is your total opposition, because XC is in vertical plane and R is in horizontal. \$\endgroup\$ – StainlessSteelRat Apr 16 '17 at 19:21
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    \$\begingroup\$ @greenhorn I = 1V/259.9 = 3.84ma. \$\endgroup\$ – Macit Apr 16 '17 at 19:22
  • \$\begingroup\$ Yes I thought that I = 1V/259.9 = 3.84mA makes more sense, but when I simulate it on the software (Proteus)... the AC ammeter gives me a circuit current of 2.72mA.... and I'm confused. \$\endgroup\$ – greenhorn Apr 16 '17 at 19:39
  • \$\begingroup\$ Is the frequency correct in your simulation? \$\endgroup\$ – StainlessSteelRat Apr 16 '17 at 20:00
  • \$\begingroup\$ Yes, 50Hz. The time is set at 100ms. \$\endgroup\$ – greenhorn Apr 16 '17 at 20:22
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Your answer to part three should be $$I=\frac{V}{Z}=\frac{1}{259.9}$$

$$P_{R_2}=I^2R_2=\frac{1}{259.9^2}50$$

The reason for this is that you have not found the complex absolute value of the impedance in your answer, (part 2 was correct, but not part 3). You need to take the square root of the product of the complex value and it's conjugate.

$$Z=(R_1+R_2)+j \left(\frac{-1}{\omega C}\right)$$ $$\vert Z\vert=\sqrt{Z^*Z }=\sqrt{(R_1+R_2)^2+\left(\frac{1}{\omega C}\right)^2}$$

If you then use Kirchoff's Voltage Law to find the 'total voltage dropped' across the circuit, you obtain the complex expression:

$$\frac{V_0 R_1+V_0R_2+V_0\frac{-j}{\omega C}}{\sqrt{(R_1+R_2)^2+\left(\frac{1}{\omega C}\right)^2}}$$

Multiplying this, by the complex conjugate and taking the square root you get

$$ \frac{ \sqrt{ V_0^2( R_1 + R_2)^2 + V_0^2\left( \frac{1}{\omega C} \right)^2 } }{\sqrt{(R_1+R_2)^2+\left(\frac{1}{\omega C}\right)^2}} =\sqrt{V_0^2}\frac{ \sqrt{ ( R_1 + R_2)^2 + \left( \frac{1}{\omega C} \right)^2 } }{\sqrt{(R_1+R_2)^2+\left(\frac{1}{\omega C}\right)^2}}=V_0$$

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