Question 19.SP.5: For a RC series circuit (Fig. 19-10), find (a) the time cons......

For a RC series circuit (Fig. 19-10), find (a) the time constant of the circuit; (b) vCv_{C} and vRv_{R} one time constant after the switch is closed and at 5 s, and (c) vCv_{C} and vRv_{R} one time constant after discharge starts, assuming the capacitor is fully charged to 10 V.

19.10
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(a) Write Eq. (19-23) for the time constant in a RC series circuit and substitute values for R and C.

T = RC      (19-23)

=(100×103)(20×106)=2 s =\left(100 \times 10^3\right)\left(20 \times 10^{-6}\right)=2\ s

(b) Write the formulas for charging voltage, substitute values, and solve for vCv_{C} and vRv_{R}.

vC=V(1et/RC)v_C=V\left(1-e^{-t / R C}\right)  (19-17)

When t = T = 2 s:

vC=10(1e2/2)=10(1e1)=10(10.368)=10(0.632)=6.32 Vv_C=10\left(1-e^{-2 / 2}\right)=10\left(1-e^{-1}\right)=10(1-0.368)=10(0.632)=6.32\ V

V=vR+vCV=v_R+v_C  (19-15)

vR=VvC=106.32=3.68 Vv_R=V-v_C=10-6.32=3.68\ V

When t = 5 s:

vC=10(1e5/2)=10(1e2.5)=10(10.082)=10(0.918)=9.18 Vv_C=10\left(1-e^{-5 / 2}\right)=10\left(1-e^{-2.5}\right)=10(1-0.082)=10(0.918)=9.18\ V

 

vR=VvC=109.18=0.82 Vv_R=V-v_C=10-9.18=0.82\ V

(c) Write the formula for discharging voltage, substitute values, and solve for vCv_{C} and vRv_{R}.

vC=Vet/RCv_C=V e^{-t / R C}  (19-22)

When t = T = 2 s:

vC=Ve2/2=Ve1=10(0.368)=3.68 Vv_C=V e^{-2 / 2}=V e^{-1}=10(0.368)=3.68\ V

0=vR+vC0=v_R+v_C  (19-20)

vR=vC=3.68 Vv_R=-v_C=-3.68\ V

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