The ABCD matrix for a two-port network is defined by- V1- I1 - A B- C D - V2- -I2 -The parameter B for the given two-port networkin ohms- correct to two decimal places is

B = .V_2I_2|_V_2=0 When V_2= V_2 nbsp; (i.e., when port 2 is short circuited). I_1 nbsp;= V_12Ω +(5Ω ||2 Ω ) = 7V_124Ω I_2 nbsp;= I_1 nbsp;× nbs ...

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Byju's Answer

Standard XII

Physics

V-I Characteristics for a P-N Junction Diode

The ABCD matr...

Question

The ABCD matrix for a two-port network is defined by
$[\begin{matrix} V_{1} I_{1} \end{matrix}] = [\begin{matrix} A & B C & D \end{matrix}] [\begin{matrix} V_{2} - I_{2} \end{matrix}]$

The parameter B for the given two-port network(in ohms, correct to two decimal places) is

4.8

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Solution

The correct option is A 4.8
$B = - {\begin{matrix} \frac{V_{2}}{I_{2}} \end{matrix} ∣ ∣ ∣}_{V_{2} = 0}$

When $V_{2} = V_{2}$ (i.e., when port-2 is short circuited).

$I_{1} = \frac{V_{1}}{2 Ω + (5 Ω | | 2 Ω)} = \frac{7 V_{1}}{24 Ω}$

$I_{2} = - I_{1} \times \frac{5 Ω}{5 Ω + 2 Ω} = \frac{- 5 V_{1}}{24 Ω}$

$S o, B = - \frac{V_{1}}{I_{2}} = \frac{24}{5} Ω = 4.80 Ω$

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The ABCD matrix for a two-port network is defined by [V1I1]=[ABCD][V2−I2] The parameter B for the given two-port network(in ohms, correct to two decimal places) is 4.8

The correct option is A 4.8B=−V2I2∣∣∣V2=0 When V2=V2 (i.e., when port-2 is short circuited). I1=V12Ω+(5Ω||2Ω)=7V124Ω I2=−I1×5Ω5Ω+2Ω=−5V124Ω So,B=−V1I2=245Ω=4.80Ω

The ABCD matrix for a two-port network is defined by
$[\begin{matrix} V_{1} I_{1} \end{matrix}] = [\begin{matrix} A & B C & D \end{matrix}] [\begin{matrix} V_{2} - I_{2} \end{matrix}]$

The parameter B for the given two-port network(in ohms, correct to two decimal places) is

4.8