Category: Network Theory Objective Questions

Network Theory Interview Questions and Answers for Experienced people on “Apparent Power and Power Factor”.

1. The highest power factor will be?
A. 1
B. 2
C. 3
D. 4
Answer: A
Clarification: The power factor is useful in determining the useful power transferred to a load. The highest power factor will be 1.

2. If power factor = 1, then the current to the load is ______ with the voltage across it.
A. out of phase
B. in phase
C. 90⁰ out of phase
D. 45⁰ out of phase
Answer: B
Clarification: If power factor = 1, then the current to the load is in phase with the voltage across it because the expression of power factor is power factor = cosθ.

3. In case of resistive load, what is the power factor?
A. 4
B. 3
C. 2
D. 1
Answer: D
Clarification: In case of resistive load, the power factor = 1 as the current to the load is in phase with the voltage across it.

4. If power factor = 0, then the current to a load is ______ with the voltage.
A. in phase
B. out of phase
C. 45⁰ out of phase
D. 90⁰ out of phase
Answer: D
Clarification: If the power factor = 0, then the current to a load is 90⁰ out of phase with the voltage and it happens in case of reactive load.

5. For reactive load, the power factor is equal to?
A. 0
B. 1
C. 2
D. 3
Answer: A
Clarification: For reactive load, the power factor is equal to 0. Power factor = 0 when current to a load is 90⁰ out of phase with the voltage.

6. Average power is also called?
A. apparent power
B. reactive power
C. true power
D. instantaneous power
Answer: C
Clarification: The average power is expressed in watts. It means the useful power transferred from the source to the load, which is also called true power. Average power is also called true power.

7. If we apply a sinusoidal voltage to a circuit, the product of voltage and current is?
A. true power
B. apparent power
C. average power
D. reactive power
Answer: B
Clarification: If we apply a sinusoidal voltage to a circuit, the product of voltage and current is apparent power. The apparent power is expressed in volt amperes or simply VA.

8. The expression of apparent power (P_app) is?
A. V_mI_m
B. V_mI_eff
C. V_effI_eff
D. V_effI_m
Answer: C
Clarification: In case of sinusoidal voltage applied to the circuit, the product of voltage and the current is not the true power or average power and it is apparent power. The expression of apparent power (P_app) is P_app = V_effI_eff.

9. The power factor=?
A. sinθ
B. cosθ
C. tanθ
D. secθ
Answer: B
Clarification: The expression of power factor is power factor= cosθ. As the phase angle between the voltage and the current increases the power factor decreases.

10. The power factor is the ratio of ________ power to the ______ power.
A. average, apparent
B. apparent, reactive
C. reactive, average
D. apparent, average
Answer: A
Clarification: The power factor is the ratio of average power to the apparent power. Power factor =(average power)/(apparent power). Power factor is also defined as the factor with which the volt amperes are to be multiplied to get true power in the circuit.

11. The power factor is called leading power factor in case of ____ circuits.
A. LC
B. RC
C. RL
D. RLC
Answer: B
Clarification: The power factor is called leading power factor in case of RC circuits and not in RLC circuits and RL circuits and LC circuits.

12. The term lagging power factor is used in which circuits?
A. RLC
B. RC
C. RL
D. LC
Answer: C
Clarification: The term lagging power factor is used in RL circuits and not in RLC circuits and RC circuits and LC circuits.

To practice all areas of Network Theory for Interviews,

Network Theory Multiple Choice Questions on “Circuit Elements in the S-Domain”.

1. The resistance element __________ while going from the time domain to frequency domain.
A. does not change
B. increases
C. decreases
D. increases exponentially

Answer: A
Clarification: The s-domain equivalent circuit of a resistor is simply resistance of R ohms that carries a current I ampere seconds and has a terminal voltage V volts-seconds. The resistance element does not change while going from the time domain to the frequency domain.

2. The relation between current and voltage in the case of inductor is?
A. v=Ldt/di
B. v=Ldi/dt
C. v=dt/di
D. v=di/dt

Answer: B
Clarification: Consider an inductor with an initial current I_o. The time domain relation between current and voltage is v=Ldi/dt.

3. The s-domain equivalent of the inductor reduces to an inductor with impedance?
A. L
B. sL
C. s²L
D. s³L

Answer: B
Clarification: If the initial energy stored in the inductor is zero, the equivalent circuit of the inductor reduces to an inductor with impedance sL ohms.

4. The voltage and current in a capacitor are related as?
A. i=Cdt/dv
B. v=Cdv/dt
C. i=Cdv/dt
D. v=Cdt/dv

Answer: C
Clarification: Consider an initially charged capacitor and the initial voltage on the capacitor is V_o. The voltage current relation in the time domain is i=Cdv/dt.

5. The s-domain equivalent of the capacitor reduces to a capacitor with impedance?
A. sC
B. C
C. 1/C
D. 1/sC

Answer: D
Clarification: The s-domain equivalent of the capacitor can be derived for the charged capacitor and it reduces to an capacitor with impedance 1/sC.

6. From the circuit shown below, find the value of current in the loop.
A. (V/R)/(s+1/RC.
B. (V/C./(s+1/R)
C. (V/C./(s+1/RC.
D. (V/R)/(s+1/R)

Answer: A
Clarification: Applying Kirchhoff’s law around the loop, we have V/s=1/sC I+RI. Solving above equation yields I=CV/(RCS+1)=(V/R)/(s+1/RC..

7. After taking the inverse transform of current in the circuit shown below, the value of current is?

A. i=(V/C.e^-t/R
B. i=(V/C.e^-t/RC
C. i=(V/R)e^-t/RC
D. i=(V/R)e^-t/R

Answer: C
Clarification: We had assumed the capacitor is initially charged to V_o volts. By taking the inverse transform of the current, we get i=(V/R) e^-t/RC.

8. The voltage across the resistor in the circuit shown below is?

A. Ve^t/R
B. Ve^-t/RC
C. Ve^-t/R
D. Ve^t/RC

Answer: B
Clarification: We can determine the voltage v by simply applying the ohm’s law from the circuit. And applying the Ohm’s law from the circuit v = Ri = Ve^-t/RC.

9. The voltage across the resistor in the parallel circuit shown is?

A. V/(s-1/R)
B. V/(s-1/RC.
C. V/(s+1/RC.
D. V/(s+1/C.

Answer: C
Clarification: The given circuit is converted to parallel equivalent circuit. By taking the node equation, we get v/R+sCv=CV. Solving the above equation, v=V/(s+1/RC..

10. Taking the inverse transform of the voltage across the resistor in the circuit shown below is?

A. Ve^-t/τ
B. Ve^t/τ
C. Ve^tτ
D. Ve^-tτ

Answer: A
Clarification: By taking the inverse transform, we get v=Ve^-t/RC=Ve^-t/τ, where τ is the time constant and τ = RC. And v is the voltage across the resistor.

Network Theory Multiple Choice Questions on “Hybrid (h) Parameter”.

1. For the circuit given below, the value of the hybrid parameter h₁₁ is ___________

A. 15 Ω
B. 20 Ω
C. 30 Ω
D. 25 Ω

Answer: A
Clarification: 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)+10right))
= 15I₁
∴ (frac{V_1}{I_1}) = 15.
Hence h₁₁ = 15 Ω.

2. For the circuit given below, the value of the hybrid parameter h₂₁ is ___________

A. 0.6 Ω
B. 0.5 Ω
C. 0.3 Ω
D. 0.2 Ω

Answer: B
Clarification: 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.

3. For the circuit given below, the value of the hybrid parameter h₁₂ is ___________

A. 6 Ω
B. 5 Ω
C. 1 Ω
D. 2 Ω

Answer: C
Clarification: 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 = I_A × 10
I_A = (frac{I_2}{2})
V₂ = I_B × 10
I_B = (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.

4. For the circuit given below, the value of the hybrid parameter h₂₂ is ___________

A. 0.2 Ω
B. 0.5 Ω
C. 0.1 Ω
D. 0.3 Ω

Answer: A
Clarification: 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 = I_A × 10
I_A = (frac{I_2}{2})
V₂ = I_B × 10
I_B = (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.

5. In the circuit given below, the value of the hybrid parameter h₁₁ is _________

A. 10 Ω
B. 7.5 Ω
C. 5 Ω
D. 2.5 Ω

Answer: B
Clarification: 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)+5right))
= 7.5I₁
∴ (frac{V_1}{I_1}) = 7.5
Hence h₁₁ = 7.5 Ω.

6. In the circuit given below, the value of the hybrid parameter h₂₁ is _________

A. 10 Ω
B. 0.5 Ω
C. 5 Ω
D. 2.5 Ω

Answer: B
Clarification: 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.

7. For the circuit given below, the value of the hybrid parameter h₁₂ is ___________

A. 6 Ω
B. 5 Ω
C. 1 Ω
D. 2 Ω

Answer: C
Clarification: 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 = I_A × 5
V₂ = I_A × 5
From the above equations, we get,
∴ (frac{V_1}{V_2} = frac{I_A×10}{I_A×10}) = 1
Hence h₁₂ = 1 Ω.

8. For the circuit given below, the value of the hybrid parameter h₂₂ is ___________

A. 0.2 Ω
B. 0.5 Ω
C. 0.1 Ω
D. 0.3 Ω

Answer: A
Clarification: 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 = I_A × 5
V₂ = I_A × 5
I_A = 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.

9. In two-port networks the parameter h₁₁ is called _________
A. Short circuit input impedance
B. Short circuit current gain
C. Open circuit reverse voltage gain
D. Open circuit output admittance

Answer: A
Clarification: 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.

10. In two-port networks the parameter h₂₁ is called _________
A. Short circuit input impedance
B. Short circuit current gain
C. Open circuit reverse voltage gain
D. Open circuit output admittance

Answer: B
Clarification: 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.

11. In two-port networks the parameter h₁₂ is called _________
A. Short circuit input impedance
B. Short circuit current gain
C. Open circuit reverse voltage gain
D. Open circuit output admittance

Answer: C
Clarification: We know that, h₂₁ = (frac{V_1}{V_2}), when I₁ = 0.
Since the current in the first loop is 0 when the ratio of the two voltages is measured, therefore the parameter h₁₂ is called as Open circuit reverse voltage gain.

12. In two-port networks the parameter h₂₂ is called _________
A. Short circuit input impedance
B. Short circuit current gain
C. Open circuit reverse voltage gain
D. Open circuit output admittance

Answer: D
Clarification: 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.

13. For a T-network if the Short circuit admittance parameters are given as y₁₁, y₂₁, y₁₂, y₂₂, then y₁₁ in terms of Hybrid parameters can be expressed as ________
A. y₁₁ = (left(- frac{h_{21} h_{12}}{h_{11}} + h_{22}right))
B. y₁₁ = (frac{h_{21}}{h_{11}})
C. y₁₁ = –(frac{h_{12}}{h_{11}})
D. y₁₁ = (frac{1}{h_{11}} )

Answer: D
Clarification: 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)).

14. For a T-network if the Short circuit admittance parameters are given as y₁₁, y₂₁, y₁₂, y₂₂, then y₁₂ in terms of Hybrid parameters can be expressed as ________
A. y₁₂ = (left(- frac{h_{21} h_{12}}{h_{11}} + h_{22}right))
B. y₁₂ = (frac{h_{21}}{h_{11}})
C. y₁₂ = –(frac{h_{12}}{h_{11}})
D. y₁₂ = (frac{1}{h_{11}} )

Answer: C
Clarification: 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)).

15. For a T-network if the Short circuit admittance parameters are given as y₁₁, y₂₁, y₁₂, y₂₂, then y₂₂ in terms of Hybrid parameters can be expressed as ________
A. y₂₂ = (left(- frac{h_{21} h_{12}}{h_{11}} + h_{22}right))
B. y₂₂ = (frac{h_{21}}{h_{11}})
C. y₂₂ = –(frac{h_{12}}{h_{11}})
D. y₂₂ = (frac{1}{h_{11}} )

Answer: A
Clarification: 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)).

Network Theory Multiple Choice Questions on “Kirchhoff’s Voltage Law”.

1. Kirchhoff’s voltage law is based on principle of conservation of ___________
A. energy
B. momentum
C. mass
D. charge
Answer: A
Clarification: KVL is based on the law of conservation of energy.

2. In a circuit with more number of loops, which law can be best suited for the analysis?
A. KCL
B. Ohm’s law
C. KVL
D. None of the mentioned
Answer: C
Clarification: KVL can be best suited for circuits with more number of loops.

3. Mathematically, Kirchhoff’s Voltage law can be _____________
A. ∑_(k=0)ⁿ(V) = 0
B. V2∑_(k=0)ⁿ(V) = 0
C. V∑_(k=0)ⁿ(V) = 0
D. None of the mentioned
Answer: A
Clarification: According to KVL, the sum of all voltages of branches in a closed loop is zero.

To practice all areas of Network Theory,

Network Theory Multiple Choice Questions 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’
Answer: B
Clarification: Millman’s Theorem states that if there are voltage sources V1, V₂,…… Vn with internal resistances R₁, R₂,…..R_n, respectively, are in parallel, then these sources are replaced by single voltage source V’ in series with R’.

2. According to Millman’s Theorem, if there are n voltage sources with n internal resistances respectively, are in parallel, then the value of equivalent voltage source is?
A. V^‘=(V₁G₁+V₂G₂+⋯.+V_nG_n)
B. V^‘=((V₁G₁+V₂G₂+⋯.+V_nG_n))/((1/G₁+1/G₂+⋯1/G_n))
C. V^‘=((V₁G₁+V₂G₂+⋯.+V_nG_n))/(G₁+G₂+⋯G_n)
D. V^‘=((V₁/G₁+V₂/G₂+⋯.+V_n/G_n))/( G₁+G₂+⋯G_n)
Answer: C
Clarification: The value of equivalent voltage source is V^‘= ((V₁G₁+V₂G₂+⋯.+V_nG_n))/(G₁+G₂+⋯G_n).

3. According to Millman’s Theorem, if there are n voltage sources with n internal resistances respectively, are in parallel, then the value of equivalent resistance is?
A. R’=G₁+G₂+⋯G_n
B. R’=1/G₁+1/G₂+⋯1/G_n
C. R’=1/((G₁+G₂+⋯G_n))
D. R’=1/(1/G₁+1/G₂+⋯1/G_n)
Answer: C
Clarification: Let the equivalent resistance is R’. The value of equivalent resistance is R’=1/((G₁+G₂+⋯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’
Answer: C
Clarification: Millman’s Theorem states that if there are current sources I₁,I₂,…… In with internal conductances G₁,G₂,…..G_n, respectively, are in series, then these sources are replaced by single current source I’ in parallel with G’.

5. According to Millman’s Theorem, if there are n current sources with n internal conductances respectively, are in series, then the value of equivalent current source is?
A. I^‘=((I₁R₁+I₂R₂+⋯.+I_nR_n))/(R₁+R₂+⋯R_n)
B. I’=I₁R₁+I₂R₂+⋯.+I_nR_n
C. I’=((I₁/R₁+I₂/R₂+⋯.+I_n/R_n))/(R₁+R₂+⋯R_n)
D. I’=I₁/R₁+I₂/R₂+⋯.+I_n/R_n
Answer: A
Clarification: The value of equivalent current source is I^‘=((I₁R₁+I₂R₂+⋯.+I_nR_n))/(R₁+R₂+⋯R_n).

6. According to Millman’s Theorem, if there are n current sources with n internal conductances respectively, are in series, then the value of equivalent conductance is?
A. G’=R₁+R₂+⋯R_n
B. G’=1/(1/R₁+1/R₂+⋯1/R_n)
C. G’=1/((R₁+R₂+⋯R_n))
D. G’=1/R₁+1/R₂+⋯1/R_n
Answer: C
Clarification: Let the equivalent conductance is G’. The value of equivalent conductance is G’=1/((R₁+R₂+⋯R_n)).

7. Calculate the current through 3Ω resistor in the circuit shown below.
millmans-theorem-q7″>
A. 1
B. 2
C. 3
D. 4
Answer: C
Clarification: 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 below using Millman’s Theorem.
millmans-theorem-q7″>millmans-theorem-q7″ alt=”network-theory-questions-answers-millmans-theorem-q7″ width=”226″ height=”106″ class=”alignnone size-full wp-image-170305″>
A. 4
B. 3
C. 2
D. 1
Answer: B
Clarification: V^‘=((V₁G₁+V₂G₂))/(G₁+G₂)=(10(1/2)+20(1/5))/(1/2+1/5)=12.86V. R’=1/((G₁+G₂))=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.
millmans-theorem-q9″>millmans-theorem-q9″ alt=”network-theory-questions-answers-millmans-theorem-q9″ width=”223″ height=”107″ class=”alignnone size-full wp-image-170307″>
A. 2
B. 1.5
C. 1
D. 0.5
Answer: B
Clarification: 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 following circuit. Find the current through 4Ω resistor using Millman’s Theorem.
millmans-theorem-q9″>millmans-theorem-q9″ alt=”network-theory-questions-answers-millmans-theorem-q9″ width=”223″ height=”107″ class=”alignnone size-full wp-image-170307″>
A. 0.5
B. 1
C. 1.5
D. 2
Answer: C
Clarification: V^‘=((V₁G₁+V₂G₂))/(G₁+G₂)=(5(1/1)+10(1/3))/(1/1+1/3)=6.25V. R’=1/((G₁+G₂))=1/(1/1+1/3)=0.75Ω. I=6.25/(4+0.75)=1.5A.