Problem 53-1
Source: Problem elaborated by the author of the site.
In the Figure 53-01.1, where V = 156.2 sin (500 t), determine:
a) the current i.
b) the voltage across the resistor and the capacitor.
c) make a phasor diagram of the circuit.
Figure 53-01.1
Solution of the Problem 53-1
Item a
From the equation of the voltage of the source that feeds the circuit, we determine the value of
ω (value that is next to t in the sine function), that is, ω = 500 rad/s.
Thus, we can calculate the capacitive reactance of the capacitor, where C = 20 x 10-6 F. Then:
XC = 1/ (ω C) = 1/ (500 x 20 x 10-6) = 100 Ω
Writing the impedance of the circuit in complex form:
Z = R + 1/ j (ω C) = 120 - j 100 Ω
Remember that in complex numbers, 1/j = -j.
We can write this impedance in polar form from the rectangular form (above) and
remembering that the impedance module is given by:
|Z| = √ (R2 + XC2) = √
(1202 + 1002) = 156.2 Ω
and the angle that the module makes with the horizontal axis, or the real numbers, is:
Therefore, the polar form for the impedance can be written as:
Z = |Z| ∠ φ = 156.2 ∠ -39.8°
With this information we can calculate the current i and the electric voltage in each component. Using polar notation, which simplifies
the calculations, we obtain:
i = 156.2 ∠ 0° / 156.2 ∠ -39.8° = 1 ∠ 39.8° A
Notice that this angle of + 39.8° shows that the electric current in the circuit is
advanced 39.8° in relation to the electric voltage applied to the circuit. We must pay attention
to the fact that the value of the voltage in the trigonometric form, that is, the maximum or peak value
(and not the rms value) was used.
Therefore, the current value is also the maximum or peak value.
If it is the desire to express the value of the electric current in the trigonometric form,
we should write:
i = 1 sin (500 t + 39.8°) A
Item b
With the value of i, we can calculate the voltage on the resistor
and the capacitor, that is:
VR = R i = 120 x 1∠ 39.8° = 120 ∠ 39.8° V
VC = XCi = 100 ∠ -90° x 1∠ 39.8° = 100 ∠ -50.2° V
Item c
Figure 53-01.2
In the Figure 53-01.2we present the phasor graph of the circuit. For reference we use the
voltage V. Then we can clearly see that the current i
is advanced of 39.8° in relation to V. Because the resistor does not cause lag in the current, the voltage on the resistor is in phase with i. We known
that the voltage in the capacitor(VC) is delayed by 90° in relation to
the electric current i.
This can be easily verified in the graphic above. Analytically we have
φ = 39.8° - (- 50.2°) = 90°.
Many students try to "prove" that the results found are correct by adding algebraically
the values of VR and VC and are surprised when they find
V = VR + VC = 120 + 100 = 220 volts. This value is completely
different from the value supplied. This is because we must sum as the voltages phasors and not
algebraically. Then, to find the correct value, it is calculated as follows:
V = √ (VR2 + VC2) = √
(1202 + 1002) = 156.2 volts
Another way of finding the same result is by calculating the sum of the horizontal components of VR and VC, that is:
