Network Elements and the Concept of Circuit
- A source Vs = 100 cos t is connected to an impedance which drawn a power Pav = 200 watts at power factor 0.8 lagging. The current drawn from the source is I = Im cos (t + θ). The value of Im and θ are given by—
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We know that
Pav = 1 Vm Im cos θ 2 200 = 1 100 Im 0.8 2
÷ Im = 5
I=Im cos (ωt + θ) = 5 cos (ωt + θ)
Since power factor is lagging and hence,
θ = – cos–1 (0.8)Correct Option: A
We know that
Pav = 1 Vm Im cos θ 2 200 = 1 100 Im 0.8 2
÷ Im = 5
I=Im cos (ωt + θ) = 5 cos (ωt + θ)
Since power factor is lagging and hence,
θ = – cos–1 (0.8)
- Pcomplex = 200 + J 150 from a source Vs = 200 cos t. The impedance across Vs is given by—
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NA
Correct Option: C
NA
- Given V2 = 2 cos 2t = 2 0°, the value of Vs is given by—
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Given that
V2 = 2 cos 2 t = 2 ∠ 0°
I L (s) = V2 = 4V2 1s / 4 s I C (s) = V2 = 3s. V2 1 / (3s / 2) 2
Total current flowing through
1 Ω = IL (s) + IC (s)V 2 = 
4 + 3s 
s 2 = V2 8 + 3s2 2s
Voltage drop across across resistance= 1. V2 8 + 3j 2 × ω2 2j × 2
On comparing with
V2 = 2 cos 2t = 2 ∠ 0º
Here, = ω = 2= 1. V2 8 – 3 × 4 2j × 2
Voltage drop across 1 Ω resistance
= V2 = V2 ∠ 90°
By using triangle law VS = 2√2 ∠ 45°Correct Option: D
Given that
V2 = 2 cos 2 t = 2 ∠ 0°I L (s) = V2 = 4V2 1s / 4 s I C (s) = V2 = 3s. V2 1 / (3s / 2) 2
Total current flowing through
1 Ω = IL (s) + IC (s)V 2 = 4 + 3s s 2 = V2 8 + 3s2 2s
Voltage drop across across resistance= 1. V2 8 + 3j 2 × ω2 2j × 2
On comparing with
V2 = 2 cos 2t = 2 ∠ 0º
Here, = ω = 2= 1. V2 8 – 3 × 4 2j × 2
Voltage drop across 1 Ω resistance
= V2 = V2 ∠ 90°
By using triangle law VS = 2√2 ∠ 45°
- In the figure Vs = 10 cos t = 10 0°. The current drawn from Vs is given by the phasor—
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For the given circuit
Z (s) = 5 || 1 = 5.(5 / s) 1 / 5(s) 5 + (5 / s) Z (s) = 25 = 5 5s + 5 s + 1 I (s) = V2 (s) = 10 ∠ 0° Z (s) 5 / s + 1
or I (s) = 2 ∠ 0° (jω + 1)
{ ∴ 10 cos t = 10 ∠ 0° ∵ ω = 1 }
= 2 ∠ 0° √2. ∠ 45°
= 2 √2 ∠ 45°Correct Option: B
For the given circuit
Z (s) = 5 || 1 = 5.(5 / s) 1 / 5(s) 5 + (5 / s) Z (s) = 25 = 5 5s + 5 s + 1 I (s) = V2 (s) = 10 ∠ 0° Z (s) 5 / s + 1
or I (s) = 2 ∠ 0° (jω + 1)
{ ∴ 10 cos t = 10 ∠ 0° ∵ ω = 1 }
= 2 ∠ 0° √2. ∠ 45°
= 2 √2 ∠ 45°
- In the circuit Vs = cos 2t. The value of C is chosen so that I from Vs is in phase with Vs. Then Ic leads IL by an angle given by—
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Given that Vs = cos 2t
I C = Vs . SC
or IC = cos ωt. s.1
or IC = cos 2t. J2 (∵ ω = 2)
or IC = 2 Vs ∠ 90° .....(1)I L = Vs = Vs = Vs = Vs 2 + Ls 2 + s 2 + jω 2 + j2 = Vs = Vs 2(1 + j) 2√2∠ 45° = Vs ∠ – 45° .....(2) 2√2
from eq. (1) and (2) it is clear that the phase difference between the Ic and IL = [90 – (– 45°)] = 135°Correct Option: C
Given that Vs = cos 2t
I C = Vs . SC
or IC = cos ωt. s.1
or IC = cos 2t. J2 (∵ ω = 2)
or IC = 2 Vs ∠ 90° .....(1)I L = Vs = Vs = Vs = Vs 2 + Ls 2 + s 2 + jω 2 + j2 = Vs = Vs 2(1 + j) 2√2∠ 45° = Vs ∠ – 45° .....(2) 2√2
from eq. (1) and (2) it is clear that the phase difference between the Ic and IL = [90 – (– 45°)] = 135°
