Network Elements and the Concept of Circuit
- The natural frequency Sn of the given circuit is—
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The circuit is redrawn to calculate the natural frequency given below.
Natural frequency is given as
Sn = – L Req
here R eq = 4 || 4 = 2L = 2 so Sn = 2 = - 1 2 Correct Option: B
The circuit is redrawn to calculate the natural frequency given below.
Natural frequency is given as Sn = – L Req
here R eq = 4 || 4 = 2L = 2 so Sn = 2 = - 1 2
- Given Vc = e– 2t (sin t + cos t) the value of VL is given by—
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VC = e– 2t (sin t + cos t)
I R = VC 1/2 = 2e– 2t (sin t + cos t) amp
I C = C d dt VC = 1. d dt e– 2t (sin t + cos t)
= e– 2t d dt sin t + sin t d dt e– 2t + e– 2t
d dt cos t + cos t d dt e– 2t
= e– 2t (cos t – sin t) + sin t (– 2) e– 2t + cos t (–2) e– 2t
= e– 2t [cos t – sin t – 2 sin t – 2 cos t]
= e– 2t [– cos t – 3 sin t] amp
so IL = IC + IR = e– 2t (– cos t – 3 sin t) + 2e– 2t (sin t + cos t)
I L = IC + IR = e– 2t (– cos t – 3 sin t) + 2e– 2t (sin t + cos t)
VL = L d dt IL = 1. d dt e– 2t (cos t – sin t)
or VL = (cos t – sin t) d dt e– 2t + e– 2t d dt (cos t – sin t)
or VL = (cos t – sin t) (– 2) e– 2t + e– 2t (– sin t – cos t)
or VL = e– 2t (sin t – 3 cos t)
or VL = e– 2t sin t – 3 e– 2t cos t volt.Correct Option: C
VC = e– 2t (sin t + cos t)
I R = VC 1/2 = 2e– 2t (sin t + cos t) amp
I C = C d dt VC = 1. d dt e– 2t (sin t + cos t)
= e– 2t d dt sin t + sin t d dt e– 2t + e– 2t
d dt cos t + cos t d dt e– 2t
= e– 2t (cos t – sin t) + sin t (– 2) e– 2t + cos t (–2) e– 2t
= e– 2t [cos t – sin t – 2 sin t – 2 cos t]
= e– 2t [– cos t – 3 sin t] amp
so IL = IC + IR = e– 2t (– cos t – 3 sin t) + 2e– 2t (sin t + cos t)
I L = IC + IR = e– 2t (– cos t – 3 sin t) + 2e– 2t (sin t + cos t)
VL = L d dt IL = 1. d dt e– 2t (cos t – sin t)
or VL = (cos t – sin t) d dt e– 2t + e– 2t d dt (cos t – sin t)
or VL = (cos t – sin t) (– 2) e– 2t + e– 2t (– sin t – cos t)
or VL = e– 2t (sin t – 3 cos t)
or VL = e– 2t sin t – 3 e– 2t cos t volt.
- Given V2 = 1 – e– 2t, the value of V1 is given by—
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Given V2 = 1 – e– 2t
I R = V2 = 3(1 – e–2t) 2 / 3 2 I c = d V2 = 1 d (1 – e– 2t) dt dt
= – (– 2) e– 2t = 2e– 2t
current across 2Ω resistance(I) = IR + IC = 3 (1 – e– 2t) + 2e– 2t 2 = 3 + 1 e– 2t 2 2 
V1 = I.2 + V2= 
3 + 1 e– 2t 
. 2 + 1 – e– 2t 2 2
= 3 + e– 2t + 1 – e– 2t
= 4Correct Option: D
Given V2 = 1 – e– 2t
I R = V2 = 3(1 – e–2t) 2 / 3 2 I c = d V2 = 1 d (1 – e– 2t) dt dt
= – (– 2) e– 2t = 2e– 2t
current across 2Ω resistance(I) = IR + IC = 3 (1 – e– 2t) + 2e– 2t 2 = 3 + 1 e– 2t 2 2
V1 = I.2 + V2= 3 + 1 e– 2t . 2 + 1 – e– 2t 2 2
= 3 + e– 2t + 1 – e– 2t
= 4
- The V-I relation for network is V = 5 – 2I, when R = 3 Ω is connected as shown the value of I is given by—
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From the relation
V = IR
5 – 2I = I.3
5 = 2I + 3I
5 = 5 I
I = 1 amp.Correct Option: D
From the relation
V = IR
5 – 2I = I.3
5 = 2I + 3I
5 = 5 I
I = 1 amp.
- For the given graphs shown below which one is nonplanar graph?
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A given graph will be non-planar if it drawn on the sheet of paper with crossing lines, otherwise it will be planar.
Figure (ii) can be treated as However the fig. (ii) cannot be drawn on the sheet of paper without crossing lines. That's why fig. (iii) is nonplanar graph. Hence alternative (D) is the correct answer.Correct Option: D
A given graph will be non-planar if it drawn on the sheet of paper with crossing lines, otherwise it will be planar.
Figure (ii) can be treated as However the fig. (ii) cannot be drawn on the sheet of paper without crossing lines. That's why fig. (iii) is nonplanar graph. Hence alternative (D) is the correct answer.
