This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Millman’s Theorem”.

1. According to Millman’s Theorem, if there are n voltage sources with n internal resistances respectively, are in parallel, then these sources are replaced by?

a) single current source I’ in series with R’

b) single voltage source V’ in series with R’

c) single current source I’ in parallel to R’

d) single voltage source V’ in parallel to R’

View Answer

Explanation: Millman’s Theorem states that if there are voltage sources V1, V

_{2},…… Vn with internal resistances R

_{1}, R

_{2},…..R

_{n}, respectively, are in parallel, then these sources are replaced by single voltage source V’ in series with R’.

2. In the question above, the value of equivalent voltage source is?

a) V^{‘}=(V_{1}G_{1}+V_{2}G_{2}+⋯.+V_{n}G_{n})

b) V^{‘}=((V_{1}G_{1}+V_{2}G_{2}+⋯.+V_{n}G_{n}))/((1/G_{1}+1/G_{2}+⋯1/G_{n}))

c) V^{‘}=((V_{1}G_{1}+V_{2}G_{2}+⋯.+V_{n}G_{n}))/(G_{1}+G_{2}+⋯G_{n})

d) V^{‘}=((V_{1}/G_{1}+V_{2}/G_{2}+⋯.+V_{n}/G_{n}))/( G_{1}+G_{2}+⋯G_{n})

View Answer

Explanation: The value of equivalent voltage source is V

^{‘}= ((V

_{1}G

_{1}+V

_{2}G

_{2}+⋯.+V

_{n}G

_{n}))/(G

_{1}+G

_{2}+⋯G

_{n}).

3. In the question above the value of equivalent resistance is?

a) R’= G_{1}+G_{2}+⋯G_{n}

b) R’=1/G_{1}+1/G_{2}+⋯1/G_{n}

c) R’=1/((G_{1}+G_{2}+⋯G_{n}) )

d) R’=1/(1/G_{1}+1/G_{2}+⋯1/G_{n})

View Answer

Explanation: Let the equivalent resistance is R’. The value of equivalent resistance is R’=1/((G

_{1}+G

_{2}+⋯G

_{n}) ).

4. According to Millman’s Theorem, if there are n current sources with n internal conductances respectively, are in series, then these sources are replaced by?

a) single voltage source V’ in parallel with G’

b) single current source I’ in series with G’

c) single current source I’ in parallel with G’

d) single voltage source V’ in series with G’

View Answer

Explanation: Millman’s Theorem states that if there are current sources I

_{1},I

_{2},…… In with internal conductances G

_{1},G

_{2},…..G

_{n}, respectively, are in series, then these sources are replaced by single current source I’ in parallel with G’.

5. In the question above, the value of equivalent current source is?

a) I^{‘}=((I_{1}R_{1}+I_{2}R_{2}+⋯.+I_{n}R_{n}))/(R_{1}+R_{2}+⋯R_{n})

b) I’= I_{1}R_{1}+I_{2}R_{2}+⋯.+I_{n}R_{n}

c) I’=((I_{1}/R_{1}+I_{2}/R_{2}+⋯.+I_{n}/R_{n}))/( R_{1}+R_{2}+⋯R_{n})

d) I’=I_{1}/R_{1}+I_{2}/R_{2}+⋯.+I_{n}/R_{n}

View Answer

Explanation: The value of equivalent current source is I

^{‘}=((I

_{1}R

_{1}+I

_{2}R

_{2}+⋯.+I

_{n}R

_{n}))/(R

_{1}+R

_{2}+⋯R

_{n}).

6. In the question above, the value of equivalent conductance is?

a) G’= R_{1}+R_{2}+⋯R_{n}

b) G’=1/(1/R_{1}+1/R_{2}+⋯1/R_{n})

c) G’=1/((R_{1}+R_{2}+⋯R_{n}) )

d) G’=1/R_{1}+1/R_{2}+⋯1/R_{n}

View Answer

Explanation: Let the equivalent conductance is G’. The value of equivalent conductance is G’=1/((R

_{1}+R

_{2}+⋯R

_{n}) ).

7. Calculate the current through 3Ω resistor in the circuit shown below.

a) 1

b) 2

c) 3

d) 4

View Answer

Explanation: Applying Nodal analysis the voltage V is given by (10-V)/2+(20-V)/5=V/3. V=8.7V. Now the current through 3Ω resistor in the circuit is I = V/3 = 8.7/3 = 2.9A ≅ 3A.

8. Find the current through 3Ω resistor in the circuit shown above using Millman’s Theorem.

a) 4

b) 3

c) 2

d) 1

View Answer

Explanation: V

^{‘}=((V

_{1}G

_{1}+V

_{2}G

_{2}))/(G

_{1}+G

_{2})=(10(1/2)+20(1/5))/(1/2+1/5)=12.86V. R’=1/((G

_{1}+G

_{2}) )=1/(1/2+1/5)=1.43Ω. Current through 3Ω resistor=I=12.86/(3+1.43)=2.9A≅3A.

9. Consider the circuit shown below. Find the current through 4Ω resistor.

a) 2

b) 1.5

c) 1

d) 0.5

View Answer

Explanation: Applying Nodal analysis the voltage V is given by (5-V)/1+(10-V)/3=V/4. V=6V. The current through 4Ω resistor I = V/4 = 6/4 = 1.5A.

10. In the circuit shown in the question 9 find the current through 4Ω resistor using Millman’s Theorem.

a) 0.5

b) 1

c) 1.5

d) 2

View Answer

Explanation: V

^{‘}=((V

_{1}G

_{1}+V

_{2}G

_{2}))/(G

_{1}+G

_{2})=(5(1/1)+10(1/3))/(1/1+1/3)=6.25V. R’=1/((G

_{1}+G

_{2}) )=1/(1/1+1/3)=0.75Ω. I=6.25/(4+0.75)=1.5A.

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