ECTION 16-4 Impedance and Admittance of Parallel RL Circuits20. What is the impedance in polar form for the circuit in Figure 16-67?21 What is the impedance in rectangular form for the circuit in Figure 16-67?22. Repeat Problem 20 for the following frequencies:(a) 1.5 kHz (b) 3 kHz (c) 5 kHz23. At what frequency does X, equal R in Figure 16-67?► FIGURE 16-6715/0° Vf=2 kHz(d) 10 kHzm 1₁L800 μεWIR12 Ω

The given circuit isWe need to determine 20. The impedance in polar form.21. The impedance in rectangular form.22. The impedance in polar form for the following frequencies a. 1.5 kHz b. 3 kHz c. 5 kHz d. 10 kHz23. The frequency at which XL=R.The reactance of the inductor is⇒XL=2πfL.The impedance of the inductor is⇒ZL=jXL=j2πfL.The impedance of the resistor is⇒ZR=R.The total impedance is the parallel combination of the impedances of inductor and resistor.⇒Z=ZL//ZR ;⇒Z=ZL.ZRZL+ZR ;⇒Z=(j2πfL)(R)j2πfL+R ;⇒Z=j2πf(800×10−6)(12)j2πf(800×10−6)+12.............................................(1)At frequency f = 2 kHz, the total impedance is⇒Z=j2π(2000)(800×10−6)(12)j2π(2000)(800×10−6)+12 ;⇒Z=j120.63712+j10.053 ;⇒Z=120.637∠90∘(12)2+(10.053)2∠tan−1(10.05312) ;⇒Z=120.637∠90∘15.654∠39.954∘ ;⇒Z=7.706∠50.045∘ Ω.The impedance in polar form for the frequency f = 2 kHz is⇒Z=7.706∠50.045∘ Ω ;By converting it into rectangular form,⇒Z=7.706[cos(50.045∘)+jsin(50.045∘)] Ω ;⇒Z=4.949+j5.907 Ω.The impedance in rectangular form for the frequency f = 2 kHz is⇒Z=4.949+j5.907 Ω.From equation (1),⇒Z=j2πf(800×10−6)(12)j2πf(800×10−6)+12 ;⇒Z=j2π(1500)(800×10−6)(12)j2π(1500)(800×10−6)+12 ;⇒Z=j90.47812+j7.539 ;⇒Z=90.478∠90∘122+7.5392∠tan−1(7.53912);⇒Z=90.478∠90∘14.172∠32.139∘ ;The impedance in rectangular form for the frequency f = 1.5 kHz is⇒Z=6.384∠57.859∘ Ω.From equation (1),⇒Z=j2πf(800×10−6)(12)j2πf(800×10−6)+12 ;⇒Z=j2π(3000)(800×10−6)(12)j2π(3000)(800×10−6)+12 ;⇒Z=j180.95612+j15.079 ;⇒Z=180.956∠90∘122+15.0792∠tan−1(15.07912);⇒Z=180.956∠90∘19.271∠51.487∘ ;The impedance in rectangular form for the frequency f = 3 kHz is⇒Z=9.389∠38.513∘ Ω.From equation (1),⇒Z=j2πf(800×10−6)(12)j2πf(800×10−6)+12 ;⇒Z=j2π(5000)(800×10−6)(12)j2π(5000)(800×10−6)+12 ;⇒Z=j301.59312+j25.133 ;⇒Z=301.593∠90∘122+25.1332∠tan−1(25.13312);⇒Z=301.593∠90∘27.851∠64.477∘ ;The impedance in rectangular form for the frequency f = 5 kHz is⇒Z=10.829∠25.523∘ Ω.From equation (1),⇒Z=j2πf(800×10−6)(12)j2πf(800×10−6)+12 ;⇒Z=j2π(10000)(800×10−6)(12)j2π(10000)(800×10−6)+12 ;⇒Z=j603.18612+j50.265 ;⇒Z=603.186∠90∘122+50.2652∠tan−1(50.26512);⇒Z=603.186∠90∘51.678∠76.573∘ ;The impedance in rectangular form for the frequency f = 10 kHz is⇒Z=11.672∠13.427∘ Ω.The given condition is⇒XL=R ;⇒2πfL=R ;⇒2πf(800×10−6)=12 ;⇒f=2387.324 Hz.The frequency at which XL=R is⇒f=2387.324 Hz.20.The impedance in polar form for the frequency f = 2 kHz is⇒Z=7.706∠50.045∘ Ω ;21.The impedance in rectangular form for the frequency f = 2 kHz is⇒Z=4.949+j5.907 Ω.22.a.The impedance in rectangular form for the frequency f = 1.5 kHz is⇒Z=6.384∠57.859∘ Ω.22.b.The impedance in rectangular form for the frequency f = 3 kHz is⇒Z=9.389∠38.513∘ Ω.22.cThe impedance in rectangular form for the frequency f = 5 kHz is⇒Z=10.829∠25.523∘ Ω.22.d.The impedance in rectangular form for the frequency f = 10 kHz is⇒Z=11.672∠13.427∘ Ω.23.The frequency at which XL=R is⇒f=2387.324 Hz.