This set of Network Theory online test focuses on “Sinusoidal Response of an R-C Circuit”.

1. In the sinusoidal response of R-C circuit, the complementary function of the solution of i is?

a) i_{c} = ce^{-t/RC}

b) i_{c} = ce^{t/RC}

c) i_{c} = ce^{-t/RC}

d) i_{c} = ce^{t/RC}

View Answer

Explanation: From the R-c circuit, we get the characteristic equation as (D+1/RC)i=-Vω/R sin(ωt+θ). The complementary function of the solution i is i

_{c}= ce

^{-t/RC}.

2. The particular current obtained from the solution of i in the sinusoidal response of R-C circuit is?

a) i_{p} = V/√(R^{2}+(1/ωC)^{2} ) cos(ωt+θ+tan^{-1}(1/ωRC))

b) i_{p}= -V/√(R^{2}+(1/ωC)^{2} ) cos(ωt+θ-tan^{-1}(1/ωRC))

c) i_{p} = V/√(R^{2}+(1/ωC)^{2} ) cos(ωt+θ-tan^{-1}(1/ωRC))

d) i_{p} = -V/√(R^{2}+(1/ωC)^{2} ) cos(ωt+θ+tan^{-1}(1/ωRC))

View Answer

Explanation: The characteristic equation consists of two parts, viz. complementary function and particular integral. The particular integral is i

_{p}= V/√(R

^{2}+(1/ωC)

^{2}) cos(ωt+θ+tan

^{-1}(1/ωRC)).

3. The value of ‘c’ in complementary function of ‘i’ is?

a) c = V/R cosθ+V/√(R^{2}+(1/(ωC))^{2} ) cos(θ+tan^{-1}(1/ωRC))

b) c = V/R cosθ+V/√(R^{2}+(1/(ωC))^{2} ) cos(θ-tan^{-1}(1/ωRC))

c) c = V/R cosθ-V/√(R^{2}+(1/(ωC))^{2} ) cos(θ-tan^{-1}(1/ωRC))

d) c = V/R cosθ-V/√(R^{2}+(1/(ωC))^{2} ) cos(θ+tan^{-1}(1/ωRC))

View Answer

Explanation: Since the capacitor does not allow sudden changes in voltages, at t = 0, i =V/R cosθ. So, c = V/R cosθ-V/√(R

^{2}+(1/(ωC))

^{2}) cos(θ+tan

^{-1}(1/ωRC)).

4. The complete solution of the current in the sinusoidal response of R-C circuit is?

a) i = e^{-t/RC}[V/R cosθ+V/√(R^{2}+(1/(ωC))^{2}) cos(θ+tan^{-1}(1/ωRC))+V/√(R^{2}+(1/ωC)^{2}) cos(ωt+θ+tan^{-1}(1/ωRC)].

b) i = e^{-t/RC}[V/R cosθ-V/√(R^{2}+(1/ωC)^{2}) cos(θ+tan^{-1}(1/ωRC))-V/√(R^{2}+(1/ωC)^{2}) cos(ωt+θ+tan^{-1}(1/ωRC)].

c) i = e^{-t/RC}[V/R cosθ+V/√(R^{2}+(1/ωC)^{2}) cos(θ+tan^{-1}(1/ωRC))-V/√(R^{2}+(1/ωC)^{2}) cos(ωt+θ+tan^{-1}(1/ωRC)].

d) i = e^{-t/RC}[V/R cosθ-V/√(R^{2}+(1/(ωC))^{2}) cos(θ+tan^{-1}(1/ωRC))+V/√(R^{2}+(1/ωC)^{2}) cos(ωt+θ+tan^{-1}(1/ωRC)].

View Answer

Explanation: The complete solution for the current becomes i = e

^{-t/RC}[V/R cosθ-V/√(R

^{2}+(1/(ωC))

^{2}) cos(θ+tan

^{-1}(1/ωRC))+V/√(R

^{2}+(1/ωC)

^{2}) cos(ωt+θ+tan

^{-1}(1/ωRC)).

5. In the circuit shown below, the switch is closed at t = 0, applied voltage is v (t) = 50cos (102t+π/4), resistance R = 10Ω and capacitance C = 1µF. The complementary function of the solution of ‘i’ is?

a) i_{c} = c exp (-t/10^{-10})

b) i_{c} = c exp(-t/10^{10})

c) i_{c} = c exp (-t/10^{-5})

d) i_{c} = c exp (-t/10^{5})

View Answer

Explanation: By applying Kirchhoff’s voltage law to the circuit, we have

(D+1/10

^{-5}) )i=-500sin(1000t+π/4). The complementary function is i

_{c}= c exp (-t/10

^{-5}).

6. The particular integral of the solution of ‘i’ from the information provided in the question 5.

a) i_{p} = (4.99×10^{-3}) cos(100t+π/4-89.94^{o})

b) i_{p} = (4.99×10^{-3}) cos(100t-π/4-89.94^{o})

c) i_{p} = (4.99×10^{-3}) cos(100t-π/4+89.94^{o})

d) i_{p} = (4.99×10^{-3}) cos(100t+π/4+89.94^{o})

View Answer

Explanation: Assuming particular integral as i

_{p}= A cos (ωt + θ) + B sin (ωt + θ)

we get i

_{p}= V/√(R

^{2}+(1/ωC)

^{2}) cos(ωt+θ-tan

^{-1}(1/ωRC))

where ω = 1000 rad/sec, θ = π/4, C = 1µF, R = 10Ω. On substituting, we get i

_{p}= (4.99×10

^{-3}) cos(100t+π/4+89.94

^{o}).

7. The current flowing in the circuit at t = 0 in the question 5 is?

a) 1.53

b) 2.53

c) 3.53

d) 4.53

View Answer

Explanation: At t = 0 that is initially current flowing through the circuit is i = V/R cosθ = (50/10)cos(π/4) = 3.53A.

8. The complete solution of ‘i’ from the information provided in the question 5.

a) i = c exp (-t/10^{-5}) – (4.99×10^{-3}) cos(100t+π/2+89.94^{o} )

b) i = c exp (-t/10^{-5}) + (4.99×10^{-3}) cos(100t+π/2+89.94^{o} )

c) i = -c exp(-t/10^{-5}) + (4.99×10^{-3}) cos(100t+π/2+89.94^{o} )

d) i = -c exp(-t/10^{-5}) – (4.99×10^{-3}) cos(100t+π/2+89.94^{o} )

View Answer

Explanation: The complete solution for the current is the sum of the complementary function and the particular integral. The complete solution for the current becomes i = c exp (-t/10

^{-5}) + (4.99×10

^{-3}) cos(100t+π/2+89.94

^{o}).

9. The value of c in the complementary function of ‘i’ in the question 5 is?

a) c = (3.53-4.99×10^{-3}) cos(π/4+89.94^{o} )

b) c = (3.53+4.99×10^{-3}) cos(π/4+89.94^{o} )

c) c = (3.53+4.99×10^{-3}) cos(π/4-89.94^{o} )

d) c = (3.53-4.99×10^{-3}) cos(π/4-89.94^{o})

View Answer

Explanation: At t = 0, the current flowing through the circuit is 3.53A. So, c = (3.53-4.99×10

^{-3}) cos(π/4+89.94

^{o}).

10. The complete solution of ‘i’ in the question 5 is?

a) i = [(3.53-4.99×10^{-3})cos(π/4+89.94^{o})] exp(-t/0.00001)+4.99×10^{-3}) cos(100t+π/2+89.94^{o} )

b) i = [(3.53+4.99×10^{-3})cos(π/4+89.94^{o})] exp(-t/0.00001)+4.99×10^{-3}) cos(100t+π/2+89.94^{o} )

c) i = [(3.53+4.99×10^{-3})cos(π/4+89.94^{o})] exp(-t/0.00001)-4.99×10^{-3}) cos(100t+π/2+89.94^{o} )

d) i = [(3.53-4.99×10^{-3})cos(π/4+89.94^{o})] exp(-t/0.00001)-4.99×10^{-3}) cos(100t+π/2+89.94^{o} )

View Answer

Explanation: The complete solution for the current is the sum of the complementary function and the particular integral. So, i = [(3.53-4.99×10

^{-3})cos(π/4+89.94

^{o})] exp(-t/0.00001)+4.99×10

^{-3}) cos(100t+π/2+89.94

^{o}).

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