Problems Resolved on R-L Circuits in AC

Problem 54-6 Source: Problem 6.10 - page 96 - EDMINISTER, Joseph A. - Book: Circuitos Elétricos - 1ª ed. - Ed. McGraw Hill - 1971

In the series-parallel circuit of the Figure 54-6.1 the effective value of the voltage between the points a-b, is 50 volts. Knowing that Z = 8.5∠30°, find the value of V.

Solution of the Problem 54-6

Denominating the impedance between the points a-b of Z_ab, we can calculate it. Like this:

Z_ab = (20 + j60) j6 / (20 + j60 + j6)

Carrying out the calculation, we find the value of Z_ab:

Z_ab = 5.502 ∠88.40° = 0.151 + j5.50 Ω

As we know V_ab and Z_ab we can calculate the current I . Let's consider V_ab, therefore V_ab = 50 ∠ 0°. So:

I = V_ab / Z_ab= 50 / 5.502∠88.40° = 9.09∠-88.40°

And as we know Z = 7.36 + j4.25 and Z_ab, we can calculate the total impedance that the circuit offers to the voltage source V, or:

Z_T = Z + Z_ab = 7.51 + j9.75 = 12.31∠52.40° Ω

Thus, V is calculated, because:

V = I Z_T = 9.09∠-88.40° x 12.31∠52.40°

Performing the calculation, the value of V is:

V = 90.49 - j65.81 = 111.89∠-36° volts

Back to Top

Previous Problem

Next Problem