Test: Series-Series Connection of Two Port Network - Electrical Engineering (EE) MCQ

In pole zero plot the two transmission zeroes are located on the jω-axis, at the complex conjugate location, and then the magnitude response exhibits a zero transmission at ω – ω_C.

Given, N = 250, L = 40 x 10^-3, d = 30 x 10^-3m, I = 160 x 10^-3A, B = 450 x 10^-3 T
Torque = 250 x 450 x 10^-3 x 40 x 10^-3 × 30 x 10^-3 x 160 x 10^-3
= 200 x 10^-6 N-m = 0.0216 N-m.

Given that, 60 V source is absorbing power, it means that current flow from positive to negative terminal in 60 V source.
Applying KVL, we get, I + I₁ = 12 A …… (1)
Current source must have the value of less than 12 A to satisfy equation (1).

Here, A = 10, B = 9, C = 11, D = 10
∴ A = D
∴ Condition for symmetry is satisfied.
Also, AD – BC = (10) (10) – (9) (11)
= 100 – 99 = 1
Therefore the condition of reciprocity is satisfied.

If the network is reciprocal, then the ratio of the response transform to the excitation transform would not vary after interchanging the position of the excitation.

In series power  = 1/p

Explanation: Given that, b = 6, n = 3
Number of Links is given by, b – n + 1
= 6 – 3 + 1 = 4.

For maximum power transfer to the load resistor RL, RL must be equal to 100 Ω.

Since, in one inductor current is leaving to dot and in other inductor current is entering to dot.
So, L_EQ = L₁ + L₂ – 2M.

In the given graph, there are 4 nodes and 6 branches.
Twig = n – 1 = 4 – 1 = 3
Link = b – n + 1 = 6 – 4 + 1 = 3.

The equivalent capacitance by applying the concept of series-parallel combination of the capacitance is,

123 c Energy delivered during talk time
E = ∫V(t)I(t) dt
Given, I (t) = 2 A = constant = 2 ∫ V(t)dt
= 2 x Shaded area

AD – BC = 1, is the condition for reciprocity for ABCD parameters, which shows that the relation is valid for reciprocal network. The ABCD parameters are obtained for the network which consists of resistance, capacitance and inductance, which indicates that it is a passive network.

The forward short circuit current gain is given by,

So, when the two networks are connected in series parallel combination,

So, h₂₁ of total network will be sum of h parameter matrices.

We know that when a Wheatstone bridge is balanced, no current will flow through the middle resistance.

The resultant R when viewed from voltage source = 100/8 = 12.5
∴ R = 12.5 – 10 || 10 = 12.5 – 5 = 7.55 Ω.