This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Attenuators”.

1. The attenuation in dB in terms of input power (P_{1}) and output power (P_{2}) is?

a) log_{10} (P_{1}/P_{2})

b)10 log_{10} (P_{1}/P_{2})

c) log_{10} (P_{2}/P_{1})

d) 10 log_{10} (P_{2}/P_{1})

View Answer

Explanation: The increase or decrease in power due to insertion or substitution of a new element in a network can be conveniently expressed in decibels or in nepers. The attenuation in dB in terms of input power (P

_{1}) and output power (P

_{2}) is Attenuation in dB = 10 log

_{10}(P

_{1}/P

_{2}).

2. If V_{1} is the voltage at port 1 and V_{2} is the voltage at port 2, then the attenuation in dB is?

a) 20 log_{10} (V_{1}/V_{2})

b) 10 log_{10} (V_{1}/V_{2})

c) 20 log_{10} (V_{2}/V_{1})

d) 10 log_{10} (V_{2}/V_{1})

View Answer

Explanation: If V

_{1}is the voltage at port 1 and V

_{2}is the voltage at port 2, then the attenuation in dB is Attenuation in dB =20 log

_{10}(V

_{1}/V

_{2}) where V

_{1}is the voltage at port 1 and V

_{2}is the voltage at port 2.

3. What is the attenuation in dB assuming I_{1} is the input current and I_{2} is the output current leaving the port?

a) 10 log_{10} (I_{1}/I_{2})

b) 10 log_{10} (I_{2}/I_{1})

c) 20 log_{10} (I_{2}/I_{1})

d) 20 log_{10} (I_{1}/I_{2})

View Answer

Explanation: Assuming I

_{1}is the input current and I

_{2}is the output current leaving the port, the attenuation in dB is Attenuation in dB =20 log

_{10}(I

_{1}/I

_{2}) where I

_{1}is the input current and I

_{2}is the output current leaving the port.

4. The value of one decibel is equal to?

a) log_{10} (N)

b) 10 log_{10} (N)

c) 20 log_{10} (N)

d) 40 log_{10} (N)

View Answer

Explanation: The value of one decibel is equal to 20 log

_{10}(N). One decibel = 20 log

_{10}(N) where N is the attenuation.

5. The value of N in dB is?

a) N= anti log (dB)

b) N= anti log(dB/10)

c) N=anti log(dB/20)

d) N=anti log(dB/40)

View Answer

Explanation: The value of N in dB can be expressed as N=anti log(dB/20).

6. In the circuit shown below, find the value of I_{1}/I_{2}.

a) (R_{1}-R_{2}+R_{0})/R_{2}

b) (R_{1}+R_{2}+R_{0})/R_{2}

c) (R_{1}-R_{2}-R_{0})/R_{2}

d) (R_{1}+R_{2}-R_{0})/R_{2}

View Answer

Explanation: R

_{2}(I

_{1}-I

_{2})=I

_{2}(R

_{1}+R

_{0})

=> I

_{2}(R

_{2}+R

_{1}+R

_{0})I

_{1}R

_{2}. On solving, I

_{1}/I

_{2}=(R

_{1}+R

_{2}+R

_{1})/R

_{2}.

7. Determine the value of N in the circuit shown in question 6.

a) (R_{1}+R_{2}-R_{0})/R_{2}

b) (R_{1}-R_{2}-R_{0})/R_{2}

c) (R_{1}+R_{2}+R_{0})/R_{2}

d) (R_{1}-R_{2}+R_{0})/R_{2}

View Answer

Explanation: N = I

_{1}/I

_{2}. We got I

_{1}/I

_{2}=(R

_{1}+R

_{2}+R

_{1})/R

_{2}. So on substituting we get N = (R

_{1}+R

_{2}+R

_{0})/R

_{2}.

8. The value of the characteristic impedance R_{0} in terms of R_{1} and R_{2} and R_{0} in the circuit shown in question 6 is?

a) R_{1}+R_{2}(R_{1}+R_{0})/(R_{1}+R_{0}+R_{2})

b) R_{1}+ R_{2}(R_{1}+R_{0})/(R_{1}+R_{0}+R_{2})

c) R_{2}+ R_{2}(R_{1}+R_{0})/(R_{1}+R_{0}+R_{2})

d) R_{0}+R_{2}(R_{1}+R_{2})/(R_{1}+R_{0}+R_{2})

View Answer

Explanation: The value of the characteristic impedance R

_{0}in terms of R

_{1}and R

_{2}and R

_{0}when it is terminated in a load of R

_{0}is R

_{0}=R

_{1}+ R

_{2}(R

_{1}+R

_{0})/(R

_{1}+R

_{0}+R

_{2}).

9. Determine the value of R_{1} in terms of R_{0} and N in the circuit shown in question 6 is?

a) R_{1}= R_{0}(N-1)/(N+1)

b) R_{1}= R_{0}(N+1)/(N+1)

c) R_{1}= R_{0}(N-1)/(N-1)

d) R_{1}= R_{0}(N+1)/(N-1)

View Answer

Explanation: R

_{0}= R

_{1}+(R

_{1}+R

_{0})/N. On solving, the value of R

_{1}in terms of R

_{0}and N is R

_{1}= R

_{0}(N-1)/(N+1).

10. Determine the value of R_{2} in terms of R_{0} and N in the circuit shown in question 6 is?

a) R_{2}= NR_{0}/(N^{2}-1)

b) R_{2}= 2 NR_{0}/(N^{2}-1)

c) R_{2}= 3 NR_{0}/(N^{2}-1)

d) R_{2}= 4 NR_{0}/(N^{2}-1)

View Answer

Explanation: NR

_{2}= R

_{1}+R

_{0}+R

_{2}. On substituting the value of R

_{1}, we get the value of R

_{2}in terms of R

_{0}and N as R

_{2}= 2 NR

_{0}/(N

^{2}-1).

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