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There are two networks connected in parallel. Thus there Yparameter will be added.
Net impedance on primary side of transformer is
Convert Δ to Y and then use series parallel combination
In the circuit given below v,n = u(—t) + u(t) V
Then vc (t) for t > 0
A voltage V = 12√2 cos 5000t V is applied to the circuit shown below.
The current i(t) is
In the circuit shown below, the voltage source has been connected for a long time enough for steady state condition to be reached. At time t = 0, switch S is opened. Then open circuit voltage across AB is
Ix = 2A
Apply KCL
4I_{x} + 2I_{x} + V_{x} = 0
V_{x} = 4I_{x} 2I_{x} = 2I_{x} = 4V
Determine RMS values of voltage across capacitor (0.25F), current throughcapacitor, if voltage across capacitor is 4 + 4 cos (9t)  5cos (4t)
Vc = 4 + 4 cos(9t)  5cos(4t)
= 1/4 {4(9) sin(9t) + 5(4) sin (4t)}
⇒ 9 sin(9t) + 5 sin (4t)
RMS value of V_{L} is =
V_{1}— V_{2} =12V................. (1)
At node 0:
(2)
At supernode 1  2
(3)
Solving 1, 2 & 3; V_{o} = 4.5V; V_{1} = 5V; V_{2} = —7V
For the circuit shown in figure, find V(t) when V_{s} = 2sin(500t) V
In the circuit shown below, the switch is closed at time t=0. The steady state value of the voltage V_{c} is_____ V
At steady state inductor short circuit, capacitor is open circuit. Thus voltage V_{c} is equal to voltage across 1 ohm resistance.
Which of the following statements are true for the circuit shown below?
1. It is first order with steady state value of
2. It is second order with steady state value of V = iV, I = 1A
3. The network function has one pole I(s)
4. The network function has two poles I(s)
Choose the correct one
Under steady state,
Where V(t) is the voltage across the capacitor
This has a pole at S = 2.
In the circuit shown below, it is found that the input ac voltage and current are in phase. The value of coupling coefficient K and the dot polarity of the coil PQ are
If input ac voltage and current are in phase then net impedance is equal to zero. Net inductive impedance = Net capacitive impedance
, Thus k = 21 and dot must be at Q so that net inductance =
The voltage applied to a particular circuit comprising two components connected in series is given by:
V = (30 + 40 sin 103t +25sin2x103t +15sin4x103t) Vand the resulting current is given by
= 0.743 sin (103t + 1.190) + 0.781sin(2x103t + 0.896) + 0.636sin(4x103t + 0.559)A
Q. Average power supplied is
Average Power P is given by
P = 0 + 5.523+ 6.099+ 4.044 =15.67W
The voltage applied to a particular circuit comprising two components connected in series is given by:
V = (30 + 40 sin103t +25sin2x103t +15sin4x103t) Vand the resulting current is given by
= 0.743 sin (103t + 1.190) + 0.781sin(2x103t + 0.896) + 0.636sin(4x103t + 0.559)A
Q. Which of the following element is not a part of the circuit?
The expression for the voltage contains a dc component of 30V. However there is no corresponding term in the expression for current. This indicates that one of the components is a capacitor (since in a dc circuit a capacitor offers infinite impedance to a direct current). Since power is delivered to the circuit the other component is a resistor.
Using current division rule
Consider the circuit shown in figure
Q. If input current is δ(t), then i_{o} (t) (in Amps) will be
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