This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “m-Derived T-Section”.

1.The relation between Z_{oT} and Z_{oT}^{‘} in the circuits shown below.

a) Z_{oT} = Z_{oT}^{‘}

b) Z_{oT} = 2 Z_{oT}^{‘}

c) Z_{oT} = 3 Z_{oT}^{‘}

d) Z_{oT} = 4 Z_{oT}^{‘}

View Answer

Explanation: The relation between Z

_{oT}and Z

_{oT}

^{’}is Z

_{oT}= Z

_{oT}

^{’}where Z

_{oT}

^{’}is the characteristic impedance of the modified (m-derived) T-network.

2. The value of Z_{2}^{’} in terms of Z_{1}, Z_{2} from the circuits shown in question 1 is?

a) Z_{2}^{‘}=Z_{2}/4 m (1-m^{2} )+Z_{2}/m

b) Z_{2}^{‘}=Z_{1}/4 m (1-m^{2} )+Z_{1}/m

c) Z_{2}^{‘}=Z_{2}/4 m (1-m^{2} )+Z_{1}/m

d) Z_{2}^{‘}=Z_{1}/4 m (1-m^{2} )+Z_{2}/m

View Answer

Explanation: As Z

_{oT}= Z

_{oT}

^{’}, √(Z

_{1}

^{2}/4+Z

_{1}Z

_{2})=√(m

^{2}Z

_{1}

^{2}/4+m Z

_{2‘}). On solving, Z

_{2}

^{‘}=Z

_{1}/(4 m (1-m

^{2}))+Z

_{2}/m.

3. The relation between Z_{oπ} and Z_{oπ}^{’} in the circuits shown below is?

a) Z_{oπ} = 2 Z_{oπ}^{’}

b) Z_{oπ} = 4 Z_{oπ}^{’}

c) Z_{oπ} = Z_{oπ}^{’}

d) Z_{oπ} = 3 Z_{oπ}^{’}

View Answer

Explanation: The characteristic impedances of the prototype and its modified sections have to be equal for matching. The relation between Z

_{oπ}and Z

_{oπ}

^{‘}is Z

_{oπ}= Z

_{oπ}

^{’}.

4. The value of Z_{1}^{‘} in terms of Z_{1}, Z_{2} from the circuits shown in question 3 is?

a) Z_{1}^{‘}=(m Z_{2}(Z_{2} 4 m)/(1-m^{2} ))/m Z_{1}(Z_{2} 4 m/(1-m^{2} ))

b) Z_{1}^{‘}=(m Z_{1}(Z_{2} 4 m)/(1-m^{2} ))/m Z_{2}(Z_{2} 4 m/(1-m^{2} ))

c) Z_{1}^{‘}=(m Z_{1}(Z_{2} 4 m)/(1-m^{2} ))/m Z_{1}(Z_{2} 4 m/(1-m^{2} ))

d) Z_{1}^{‘}=(m Z_{1}(Z_{2} 4 m)/(1-m^{2} ))/m Z_{1}(Z_{1} 4 m/(1-m^{2} ))

View Answer

Explanation: As Z

_{oπ}= Z

_{oπ}

^{’}, √(Z

_{1}Z

_{2}/(1+Z

_{1}/4 Z

_{2}))=√(((Z

_{1}

^{‘}Z

_{2})/m)/(1+(Z

_{1}

^{‘})/(4 Z

_{2}/m))). On solving, Z

_{1}

^{‘}=(m Z

_{1}(Z

_{2}4 m)/(1-m

^{2}))/m Z

_{1}(Z

_{2}4 m/(1-m

^{2})) .

5. The value of resonant frequency in the m-derived low pass filter is?

a) f_{r}=1/(√(LC(1+m^{2} ) ))

b) f_{r}=1/(√(πLC(1+m^{2} ) ))

c) f_{r}=1/(√(LC(1-m^{2} ) ))

d) f_{r}=1/(√(πLC(1-m^{2} ) ))

View Answer

Explanation: ω

_{r}

^{2}= 1/(LC(1-m

^{2})). So the value of resonant frequency in the m-derived low pass filter is f

_{r}=1/√(πLC(1-m

^{2}) ).

6. The cut-off frequency of the low pass filter is?

a) 1/√LC

b) 1/(π√LC)

c) 1/√L

d) 1/(π√L)

View Answer

Explanation: To determine the cut-off frequency of the low pass filter we place m = 0. So f

_{c}=1/(π√LC).

7. The resonant frequency of m-derived low pass filter in terms of the cut-off frequency of low pass filter is?

a) f_{c}/√(1-m^{2} )

b) f_{c}/√(1+m^{2} )

c) f_{c}/(π√(1-m^{2} ))

d) f_{c}/(π√(1+m^{2} ))

View Answer

Explanation: If a sharp cut-off is desired, the frequency at infinity should be near to f

_{c}. The resonant frequency of m-derived low pass filter in terms of the cut-off frequency of low pass filter is f

_{r}=f

_{c}/√(1-m

^{2}).

8. The expression of m of the m-derived low pass filter is?

a) m=√(1+(f_{c}/f_{r})^{2} )

b) m=√(1+(f_{c}/f)^{2})

c) m=√(1-(f_{c}/f_{r})^{2} )

d) m=√(1-(f_{c}/f)^{2} )

View Answer

Explanation: As f

_{r}=f

_{c}/√(1-m

^{2}). The expression of m of the m-derived low pass filter is m=√(1-(f

_{c}/f

_{r})

^{2}).

9. Given a m-derived low pass filter has cut-off frequency 1 kHz, design impedance of 400Ω and the resonant frequency of 1100 Hz. Find the value of k.

a) 400

b) 1000

c) 1100

d) 2100

View Answer

Explanation: The value of k is equal to the design impedance. Given design impedance is 400Ω. So, k = 400.

10. The value of m from the information provided in question 9.

a) 0.216

b) 0.316

c) 0.416

d) 0.516

View Answer

Explanation: m=√(1-(f

_{c}/f

_{r})

^{2}) f

_{c}= 1000, f

_{r}= 1100. On substituting m=√(1-(1000/1100)

^{2})=0.416.

**Sanfoundry Global Education & Learning Series – Network Theory.**

To practice all areas of Network Theory, __here is complete set of 1000+ Multiple Choice Questions and Answers__.