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The correct answer is (a) 7.5 Ω

For explanation: Inverse HYBRID parameter G22 is given by, g22 = \(\frac{V_2}{I_2}\), when V1 = 0.

Therefore short circuiting the TERMINAL X-X’ we get,

-5 I2 – 5 I1 + V2 = 0

-5 I1 – 5(I1 – I2) = 0

Or, 2I1 = I2

Putting the above equation in the first equation, we get,

-7.5 I2 = -V2

Or, \(\frac{V_2}{I_2}\)= 7.5

Hence g22 = 7.5 Ω.

Correct answer is (d) h11 h22 – h12 h22 = 1

For EXPLANATION: We know that, I1 = y11 V1 + y12 V2 ……… (1)

I2 = y21 V1 + Y22 V2 ………. (2)

And, V1 = h11 I1 + h12 V2 ………. (3)

I2 = h21 I1 + h22 V2 ……….. (4)

Now, (3) and (4) can be rewritten as,

I1 = \(\frac{V_1}{h_{11}}– \frac{h_{12} V_2}{h_{11}}\)………. (5)

And I2 = \(\frac{h_{21} V_1}{h_{11}}+ \left(- \frac{h_{21} h_{12}}{h_{11}} + h_{22}\right) V_2\) ………. (6)

Therefore using the above 6 equations in representing the hybrid parameters in terms of the Y parameters and applying ∆Y=0, we GET,

h11 h22 – h12 h22 = 1 [hence proved].

Right choice is (b) 0.133 Ω

The BEST explanation: Inverse Hybrid parameter G11 is given by, g11 = \(\frac{I_1}{V_1}\), when I2 = 0.

Therefore SHORT circuiting the terminal Y-Y’, we get,

 V1 = I1 ((5 || 5) + 5)

= I1 \(\LEFT(\left(\frac{5×5}{5+5}\right)+5\right)\)

= 7.5I1

∴ \(\frac{I_1}{V_1} = \frac{1}{7.5}\) = 0.133

Hence g11 = 7.5 Ω.

Right choice is (b) 0.5 Ω

Best explanation: INVERSE Hybrid parameter G21 is given by, g21 = \(\frac{V_2}{V_1}\), when I2 = 0.

Therefore short circuiting the TERMINAL Y-Y’, and APPLYING KIRCHHOFF’s law, we get,

V1 = I1 (5 + 5)

V2 = I1 5

∴ \(\frac{V_2}{V_1}= \frac{I_1 5}{I_1 10}\) = 0.5

 Hence g21 = 0.5 Ω.

Right ANSWER is (c) Z

Explanation: Y-parameter = \(\frac{1}{Z}\)[1; -1; -1; 1]

And from the DEFINITION of the Y PARAMETERS, ∆Y = 0. Therefore the Y-parameter exists.

Since ∆Y = 0, so by property of reciprocity, ∆h = 0 and ∆g = 0.

Hence both hybrid and inverse hybrid parameters exist.

But the Z-parameters cannot exist here because if one terminal is opened the circuit will become invalid.

∴ Z- parameters do not exists.

The correct OPTION is (d) [Z1; Z1; Z1; Z1 + Z2]

To explain I WOULD say: z11 = \(\FRAC{V_1}{I_1}\), when I2 = 0

z22 = \(\frac{V_2}{I_2}\), when I1 = 0

z12 = \(\frac{V_1}{I_2}\), when I1 = 0

z21 = \(\frac{V_2}{I_1}\), when I2 = 0

Now, in the given CIRCUIT putting I1 = 0, we get,

z12 = Z1 and z22 = Z1 + Z2

And putting I2 = 0, we get,

z21 = Z1 and z11 = Z1.

Correct choice is (a) y22 = \(\left(- \FRAC{h_{21} h_{12}}{h_{11}} + h_{22}\RIGHT)\)

The best explanation: We know that the short circuit admittance parameters can be expressed in terms of voltages and currents as,

I1 = y11 V1 + Y12 V2 ……… (1)

I2 = y21 V1 + y22 V2 ………. (2)

And the Hybrid parameters can be expressed in terms of voltages and currents as,

V1 = H11 I1 + h12 V2 ………. (3)

I2 = h21 I1 + h22 V2 ……….. (4)

Now, (3) and (4) can be rewritten as,

I1 = \(\frac{V_1}{h_{11}}– \frac{h_{12} V_2}{h_{11}}\)………. (5)

And I2 = \(\frac{h_{21} V_1}{h_{11}}+ \left(- \frac{h_{21} h_{12}}{h_{11}} + h_{22}\right) V_2\) ………. (6)

∴ Comparing (1), (2) and (5), (6), we GET,

y11 = \(\frac{1}{h_{11}} \)

y12 = –\(\frac{h_{12}}{h_{11}}\)

y21 = \(\frac{h_{21}}{h_{11}}\)

y22 = \(\left(- \frac{h_{21} h_{12}}{h_{11}} + h_{22}\right)\).

Correct ANSWER is (B) 0.5 Ω

The best explanation: Inverse Hybrid parameter g21 is given by, g21 = \(\frac{V_2}{V_1}\), when I2 = 0.

Therefore short circuiting the terminal Y-Y’, and applying Kirchhoff’s law, we get,

V1 = I1 (10 + 10)

V2 = I1 10

∴\(\frac{V_2}{V_1}= \frac{I_1 10}{I_1 20}\) = 0.5

 Hence g21 = 0.5 Ω.

Right answer is (a) 0.067 Ω

The EXPLANATION is: INVERSE Hybrid parameter g11 is given by, g11 = \(\frac{I_1}{V_1}\), when I2 = 0.

Therefore SHORT circuiting the terminal Y-Y’, we get,

 V1 = I1 ((10||10) + 10)

= I1 \(\left(\left(\frac{10×10}{10+10}\right)+10\right)\)

= 15I1

∴ \(\frac{I_1}{V_1} = \frac{1}{15}\) = 0.067 Ω

Hence g11 = 15 Ω.

The correct option is (c) y12 =–\(\frac{h_{12}}{h_{11}}\)

Explanation: We know that the short circuit admittance parameters can be expressed in terms of voltages and currents as,

I1 = y11 V1 + y12 V2 ……… (1)

I2 = y21 V1 + y22 V2 ………. (2)

And the Hybrid parameters can be expressed in terms of voltages and currents as,

V1 = H11 I1 + h12 V2 ………. (3)

I2 = h21 I1 + h22 V2 ……….. (4)

Now, (3) and (4) can be rewritten as,

I1 = \(\frac{V_1}{h_{11}}– \frac{h_{12} V_2}{h_{11}}\)………. (5)

And I2 = \(\frac{h_{21} V_1}{h_{11}}+ \left(- \frac{h_{21} h_{12}}{h_{11}} + h_{22}\right) V_2\) ………. (6)

∴ Comparing (1), (2) and (5), (6), we GET,

y11 = \(\frac{1}{h_{11}} \)

y12 = –\(\frac{h_{12}}{h_{11}}\)

y21 = \(\frac{h_{21}}{h_{11}}\)

y22 = \(\left(- \frac{h_{21} h_{12}}{h_{11}} + h_{22}\right)\).

The correct answer is (d) y11 = \(\frac{1}{h_{11}} \)

Easy explanation: We know that the short circuit admittance PARAMETERS can be expressed in terms of voltages and currents as,

I1 = y11 V1 + y12 V2 ……… (1)

I2 = y21 V1 + y22 V2 ………. (2)

And the Hybrid parameters can be expressed in terms of voltages and currents as,

V1 = h11 I1 + h12 V2 ………. (3)

I2 = h21 I1 + h22 V2 ……….. (4)

Now, (3) and (4) can be REWRITTEN as,

I1 = \(\frac{V_1}{h_{11}}– \frac{h_{12} V_2}{h_{11}}\)………. (5)

And I2 = \(\frac{h_{21} V_1}{h_{11}}+ \left(- \frac{h_{21} h_{12}}{h_{11}} + h_{22}\RIGHT) V_2\) ………. (6)

∴ Comparing (1), (2) and (5), (6), we get,

y11 = \(\frac{1}{h_{11}} \)

y12 = –\(\frac{h_{12}}{h_{11}}\)

y21 = \(\frac{h_{21}}{h_{11}}\)

y22 = \(\left(- \frac{h_{21} h_{12}}{h_{11}} + h_{22}\right)\).

Correct option is (d) Open circuit OUTPUT admittance

The EXPLANATION is: We know that, h22 = \(\frac{I_2}{V_2}\), when I1 = 0.

Since the CURRENT in the first LOOP is 0 when the ratio of the current and voltage in the second loop is measured, therefore the parameter h22 is called as Open circuit output admittance.

The correct choice is (B) Short circuit current gain

Best explanation: We know that, h21 = \(\FRAC{I_2}{I_1}\), when V2 = 0.

Since the second output terminal is short circuited when the RATIO of the TWO currents is measured, therefore the parameter h21 is called Short circuit current gain.

Correct choice is (a) SHORT circuit INPUT impedance

Explanation: We know that, h11 = \(\frac{V_1}{I_1}\), when V2 = 0.

Since the SECOND output terminal is short circuited when the ratio of the TWO voltages is measured, therefore the parameter h11 is called as Short circuit input impedance.

Correct choice is (a) 0.2 Ω

The EXPLANATION: Hybrid parameter h22 is given by, h22 = \(\FRAC{I_2}{V_2}\), when I1 = 0.

Therefore short CIRCUITING the TERMINAL X-X’ we get,

V1 = IA × 5

V2 = IA × 5

IA = I2

From the above equations, we get,

∴ \(\frac{I_2}{V_2} = \frac{I_2}{I_2×5}\) = 0.2

Hence h22 = 0.2 Ω.

The correct choice is (c) 1 Ω

For EXPLANATION: HYBRID parameter H12 is given by, h12 = \(\frac{V_1}{V_2}\), when I1 = 0.

Therefore short circuiting the terminal X-X’ we get,

V1 = IA × 5

V2 = IA × 5

From the above equations, we get,

∴ \(\frac{V_1}{V_2} = \frac{I_A×10}{I_A×10}\) = 1

Hence h12 = 1 Ω.

Correct CHOICE is (b) 0.5 Ω

Explanation: Hybrid parameter h21 is GIVEN by, h21 = \(\frac{I_2}{I_1}\), when V2 = 0.

Therefore SHORT circuiting the terminal Y-Y’, and applying Kirchhoff’s LAW, we get,

-5 I2 – (I2 – I1)5 = 0

Or, -I2 = I2 – I1

Or, -2I2 = -I1

∴\(\frac{I_2}{I_1} = \frac{1}{2}\)

 Hence h21 = 0.5 Ω.

Correct answer is (a) 0.2 Ω

To explain I would SAY: Hybrid parameter H22 is given by, h22 = \(\frac{I_2}{V_2}\), when I1 = 0.

Therefore short CIRCUITING the terminal X-X’ we get,

V1 = IA × 10

IA = \(\frac{I_2}{2}\)

V2 = IB × 10

IB = \(\frac{I_2}{2}\)

From the above 4 equations, we get,

∴ \(\frac{I_2}{V_2}= \frac{I_2×2}{I_2×10}\) = 0.2

Hence h22 = 0.2 Ω.

Correct choice is (b) 7.5 Ω

To explain I would say: HYBRID parameter h11 is given by, h11 = \(\frac{V_1}{I_1}\), when V2 = 0.

Therefore short circuiting the terminal Y-Y’, we get,

 V1 = I1 ((5 || 5) + 5)

= I1 \(\LEFT(\left(\frac{5×5}{5+5}\right)+5\right)\)

= 7.5I1

∴ \(\frac{V_1}{I_1}\) = 7.5

Hence h11 = 7.5 Ω.

Right choice is (C) 1 Ω

The best I can explain: HYBRID PARAMETER h12 is GIVEN by, h12 = \(\frac{V_1}{V_2}\), when I1 = 0.

Therefore short circuiting the terminal X-X’ we get,

V1 = IA × 10

IA = \(\frac{I_2}{2}\)

V2 = IB × 10

IB = \(\frac{I_2}{2}\)

From the above 4 EQUATIONS, we get,

∴ \(\frac{V_1}{V_2} = \frac{I_2×10}{I_2×10}\) = 1

Hence h12 = 1 Ω.

Correct option is (b) 0.5 Ω

Best EXPLANATION: Hybrid parameter h21 is given by, h21 = \(\frac{I_2}{I_1}\), when V2=0.

Therefore SHORT CIRCUITING the terminal Y-Y’, and applying Kirchhoff’s LAW, we get,

-10 I2 – (I2 – I1)10 = 0

Or, -I2 = I2 – I1

Or, -2I2 = -I1

∴\(\frac{I_2}{I_1} = \frac{1}{2}\)

 Hence h21 = 0.5 Ω.

Right choice is (a) 15 Ω

To elaborate: Hybrid parameter h11 is given by, h11 = \(\frac{V_1}{I_1}\), when V2=0.

Therefore SHORT circuiting the terminal Y-Y’, we get,

 V1 = I1 ((10||10) + 10)

= I1 \(\LEFT(\left(\frac{10×10}{10+10}\right)+10\right)\)

= 15I1

∴ \(\frac{V_1}{I_1}\) = 15.

Hence h11 = 15 Ω.

Correct answer is (a) 1/2

Explanation: Open CIRCUITING a-a^‘ we GET V1=Iy×2 and

Iy=I2/2 and V2=Ix×4 and Ix=I2/2. On solving and SUBSTITUTING, we get H12 = V1/V2=1/2.

Correct option is (b) 2

The explanation: h11=V1/I1 |V2=0. So short CIRCUITING b-b^‘, V1 = I1((2||2)+1) = 2I1 => V1/I1 = 2. On substituting we get h11 = V1/I1 = 2Ω.

The CORRECT ANSWER is (C) short circuit forward current gain

Easy explanation: h21=I2/I1 |V2=0. So the hybrid PARAMETER h21 is called short circuit forward current gain.

Correct CHOICE is (a) short circuit INPUT IMPEDANCE

Best EXPLANATION: h11=V1/I1 |V2=0. So the hybrid parameter h11 is called short circuit input impedance.

The correct OPTION is (C) 7/5

For explanation I WOULD say: D is a transmission parameter and is given by D = -I1/I2 |V2=0. Short circuiting b-b^‘, I1 = 7/17 V1 and-I2 = 5/17 V1. So we get I1/I2 = 7/5. So D=7/5.

The correct choice is (b) 17/5

To explain: The transmission parameter B is GIVEN by B = -V1/I2|V2=0. SHORT circuiting b-b^‘, -I2 = 5/17 V1 => -V1/I2 = 17/5. On substituting we get B=17/5 Ω.

The correct CHOICE is (d) 1/5

For explanation I would SAY: C = I1/V2|I2=0. By open circuiting b-b^‘ we GET V2 = 5 I1 => I1/V2 = 1/5. On substituting we get C = I1/V2=1/5 Ω.

The correct OPTION is (a) 6/5

The explanation is: OPEN circuiting b-b^‘, V1 = 6 I1, V2 = 5I1. On solving V1/V2 = 6/5. On substituting we get A = V1/V2=6/5.

Right OPTION is (d) I1=0.5(V1)-0.25(V2)

The best I can EXPLAIN: We got the admittance parameters as Y11 = 0.5, Y12 = -0.25, Y21 = -0.25, Y22 = 0.625. So the equations in TERMS of admittance parameters is I1=0.5(V1)-0.25(V2) and I2=-0.25(V1)+0.625(V2).

The CORRECT answer is (c) -1/4

Easy explanation: Short circuiting a-a’, -I1 = 2/5 I2 and I2 = 5 V2/8. On solving -I1 = 2/5×5/8 V2 = V2/4. We know Y12 = I1/V2. We GOT I1/V2 = -1/4. So the value of Y12 will be -1/4 mho.

The correct OPTION is (b) 5/8

To elaborate: On SHORT circuiting a-a’, we GET ZEQ = 8/5 Ω. V2=I2× 8/5. We know Y22 = I2/V2. We got I2/V1 = 5/8. ON SUBSTITUTING we get Y22 = 5/8 mho.

The correct option is (a) -1/4

The EXPLANATION is: After SHORT circuiting b-b’, the equation will be -I2=I1 × 2/4=I1/2 and -I2 = V1/4 and on SOLVING and substituting we GET Y21 = I2/V1=-1/4 mho.

Correct ANSWER is (d) 1/2

To EXPLAIN I would say: After SHORT circuiting b-b’, the EQUATION will be V1 = (I1) x 2. We know Y11 = I1/V1. From the equation we get I1/V1 = 2. On SUBSTITUTING we get Y11 = 2 mho.

Right ANSWER is (B) Short circuit admittance parameters

To explain I would say: The parameters Y11, Y12, Y21, Y22 are called short circuit admittance parameters also called network functions as they are obtained by short circuiting PORT 1 or port 2.

Right answer is (C) I2 = Y21V1 + Y22 V2

The best I can explain: The expression RELATING the voltages V1, V2 and current is I2 and short CIRCUIT parameters Y11, Y12 is I2 = Y21V1 + Y22V2.

Correct choice is (a) I1 = Y11 V1 + Y12 V2

The best I can EXPLAIN: The expression RELATING the short circuit parameters Y11, Y12 andvoltages V1, V2 and CURRENT is I1, is I1 = Y11 V1 + Y12 V2.

The correct option is (d) V1 and V2

The best I can explain: In determining short CIRCUIT impedance parameters, among V1, V2, I1, I2; V1 and V2 are independent VARIABLES and I1, I2 are dependent variables. Independent variables are the variables that do not depend on any other VARIABLE.

The correct answer is (b) I1 and I2

The explanation is: In determining short circuit impedance parameters, among V1, V2, I1, I2; I1 and I2 are dependent VARIABLES and V1, V2 are independent variables i.e., dependent variables depend on independent variables.

The correct choice is (c) 1

Explanation: The CURRENT through VERTICAL 2Ω RESISTOR is = I2/2. So, V1 = 2 x (I2/2). On solving and substituting we get Z12 = 1Ω.

The correct choice is (c) 2

For explanation I WOULD SAY: Open circuiting port 1, we get V2 = I2((2+2)||4) => V2 = I2 x 2 => V2/I2 = 2. Therefore the VALUE of Z22 is 2Ω.

Right answer is (B) 1

Easy explanation: V2 is the voltage across the 4Ω impedance. The CURRENT through 4Ω impedance is I1/4. And V2 = (I1/4) x 4 = I1. So, Z21 = 1Ω.

The correct CHOICE is (d) 2.5

The BEST explanation: For DETERMINING Z11, the current I2 is equal to zero. Now we obtain Zeq as 1 + (6×2)/(6+2)=2.5Ω. So, Z11 = 2.5Ω.

The correct CHOICE is (b) V2 = Z21I1 + Z22 I2

The explanation: The expression relating the currents I1, I2 and voltage V1 and OPEN circuit parameters Z21, Z22 is V2 = Z21I1 + Z22 I2.

The correct option is (c) I1 and I2

The best I can EXPLAIN: In determining open circuit IMPEDANCE parameters, among V1, V2, I1, I2; I1 and I2 are independent variables and V1, V2 are dependent variables. Independent variables are the variables that do not depend on any other variable.

Correct OPTION is (c) V1 = Z11I1 + Z12 I2

To EXPLAIN: The expression relating the open CIRCUIT parameters Z11, Z12 andcurrents I1, I2 and VOLTAGE V1 is V1 = Z11I1 + Z12 I2.