300+ Network Theorem MCQ- Objective Question Answer for Network Theorem Quiz

136. In the figure given below, the Norton Resistance, as seen at the terminals P − Q, is given by __________

A. 5 Ω
B. 7.5 Ω
C. 5 Ω
D. 7.5 Ω

Answer: A

For finding V_N,

V_N = (4 × 10)/(10 + 10) = 2V

For finding R_N,

R_N = 10 || 10

= (10×10)/(10 + 10) = 5 Ω.

137. The Norton equivalent impedance Z between the nodes P and Q in the following circuit is __________

A. 1
B. 1 + s + 1/s
C. 2 + s + 1/s
D. 3 + s + 1/s

Answer: A

To calculate the Norton resistance, all the current sources get open − circuited and voltage sources get short − circuited.

∴ R_N = (1/s + 1) || (1 + s)

$\frac{{\left( {\frac{1}{s} + 1} \right) \times (1 + s)}}{{\left( {\frac{1}{s} + 1} \right) + (1 + s)}}$

$\frac{{\frac{1}{s} + 1 + 1 + s}}{{\frac{1}{s} + 1 + 1 + s}}$

So, R_N = 1.

138. In the circuit given below, it is given that V_AB = 4 V for R_L = 10 kΩ and V_AB = 1 V for R_L = 2kΩ. The value of Norton resistance for the network N is ________

A. 16 kΩ
B. 30 kΩ
C. 3 kΩ
D. 50 kΩ

Answer: B

When R_L = 10 kΩ and V_AB = 4 V

Current in the circuit

V_AB/R_L = 4/10 = 0.4 mA

Norton voltage is given by

V_N = I (R_N + R_L)

= 0.4(R_N + 10)

= 0.4R_N + 4

Similarly, for R_L = 2 kΩ and V_AB = 1 V

So, I = 1/2 = 0.5 mA

V_N = 0.5(R_N + 2)

= 0.5 R_N + 1

∴ 0.1R_N = 3

Or, R_N = 30 kΩ.

139. For the circuit given below, Norton’s resistance across the terminals A and B is _________

A. 5 Ω
B. 7 kΩ
C. 1.5 kΩ
D. 1.1 kΩ

Answer: B

Let V_AB = 1 V

5 V_AB = 5

Or, 1 = 1 × I₁ or, I₁ = 1

Also, 1 = − 5 + 1(I – I₁)

∴ I = 7

Hence, R = 7 kΩ.

140. A circuit is given in the figure below. The Norton equivalent as viewed from terminals x and x’s is ___________

A. 6 Ω and 1.333 A
B. 6 Ω and 0.833 A
C. 32 Ω and 0.156 A
D. 32 Ω and 0.25 A

Answer: B

We, draw the Norton equivalent of the left side of xx’ and source transformed right side of yy’.

V_xx’ = V_N = $\displaystyle\frac{\frac{4}{8} + \frac{8}{24}}{\frac{1}{8} + \frac{1}{24}}$ = 5V

∴ R_N = 8 || (16 + 8)

= (8×24)/(8 + 24) = 6 Ω

∴ (I_N = V_N/R_N = 5/6 = 0.833 A.

141. For the circuit given in the figure below, the Norton equivalent as viewed from terminals y and y’ is _________

A. 32 Ω and 0.25 A
B. 32 Ω and 0.125 A
C. 6 Ω and 0.833 A
D. 6 Ω and 1.167 A

Answer: D

We draw the Norton equivalent of the left side of xx’ and source transformed right side of yy’.Norton equivalent as seen from terminal yy’ is

Vxx’ = VN = $\displaystyle\frac{\frac{4}{8} + \frac{8}{24}}{\frac{1}{8} + \frac{1}{24}}$ = 5V

= (0.167 + 1)/(0.04167 + 0.125) = 7 V
∴ RN = (8 + 16) || 8
= (24×8)/(24 + 8) = 6 Ω

∴I_N = V_N/R_N = 7/6 = 1.167 A.

142. In the figure given below, the power loss in 1 Ω resistor using Norton’s Theorem is ________

A. 9.76 W
B. 9.26 W
C. 10.76 W
D. 11.70 W

Answer: B

Let us remove the 1 Ω resistor and short x − y.
At Node 1, assuming node potential to be V,

(V − 10)/5 + I_SC = 5

But I_SC = V/2

∴ (V − 10)/5 + V/2 = 5

Or, 0.7 V = 7

That is, V = 10 V

∴ I_SC = V/2 = 5 A

To find R_int, all constant sources are deactivated.

R_int = (5 + 2)×2/(5 + 2 + 2) = 14/9 = 1.56 Ω

R_int = 1.56 Ω; I_SC = I_N = 5A

Here, I = I_N

$\frac{R_{int}}{R_{int} + 1} = 5 × \frac{1.56}{1.56 + 1}$ = 3.04 A

∴ Power loss = (3.04)² × 1 = 9.26 W.

143. The value of R_N from the circuit given below is ________

A. 3 Ω
B. 1.2 Ω
C. 5 Ω
D. 12.12 Ω

Answer: D

V_X = 3V_X/6 + 4

V_X = 8 V = V_OC

If terminal is short − circuited, V_X = 0.

I_SC = 4/3 + 3) = 0.66 A

∴ R_N = V_OC/I_SC = 8/0.66 = 12.12 Ω.

144. By using the Norton equivalent theorem find the current I, as shown in the figure below, is ________

A. 3 A
B. 2 A
C. 1 A
D. 0

Answer: C

The 3 Ω resistance is an extra element because the voltage at node B is independent of the 3 Ω resistance.

I₁ = 3/2 + 1 = 1 A (B − > A.

The net current in 2 Ω resistance is

I = 1 – I₁

= 2 – 1 = 1 A (A − > B..

145. While computing the Norton equivalent resistance and the Norton equivalent current, which of the following steps are undertaken?

Both the dependent and independent voltage sources are short-circuited and both the dependent and independent current sources are open-circuited
Both the dependent and independent voltage sources are open-circuited and both the dependent and independent current sources are short-circuited
The dependent voltage source is open-circuited keeping the independent voltage source untouched and the dependent current source is short-circuited keeping the independent current source untouched
The dependent voltage source is short-circuited keeping the independent voltage source untouched and the dependent current source is open-circuited keeping the independent current source untouched

Answer: D
While computing the Norton equivalent voltage consisting of both dependent and independent sources, we first find the equivalent resistance called the Norton resistance by opening the two terminals.

Then while computing the Norton current, we short-circuit the dependent voltage sources keeping the independent voltage sources untouched, and open-circuiting the dependent current sources keeping the independent current sources untouched.

141. A voltage source connected in series with a resistor can be converted to a?

A. Current source in series with a resistor
B. Current source in parallel with a resistor
C. Voltage source in parallel with a resistor
D. Cannot be modified

Answer: B

A voltage source connected in series can be converted to a current source connected in parallel using the relation obtained from Ohm’s law, that is

V = IR.

This equation shows that a voltage source connected in series has the same impact as a current source connected in parallel.

142. Calculate the total current in the circuit by using source Transformation.

A. 2.3mA
B. 4.3mA
C. 3.3mA
D. 1.3mA

Answer: C

The 9mA source connected in parallel to the 5 Kohm resistor can be converted to a 45V source in series with a 5 Kohm resistor. Applying mesh analysis, we get:

I = (45 − 3)/(5 + 4.7 + 3) = 3.3mA.

143. Find the value of voltage once source transformation is applied to the circuit.

A. 10V
B. 30V
C. 50V
D. 70V

Answer: C

Using ohm’s law, we can use the relation:

V = IR.

Thus V = 10 × 5 = 50V.

144. Once the circuit is transformed into a voltage source where will the resistance be connected?

A. In series with the voltage source
B. In parallel with the voltage source
C. The resistance is removed from the circuit
D. Resistance is multiplied by 10 and connected in series with the source

Answer: A

The resistance is connected in series with the voltage source because we are transforming a current source connected in parallel to a resistor into a voltage source connected in series with it.

145. What will the value of the current be once source transformation is applied to the circuit?

A. 10A
B. 20A
C. 30A
D. 40A

Answer: A

Using ohm’s law, we can use the relation:

V = IR.

Thus I = V/R.

I = 220/22 = 10A.

146. Once the circuit is transformed into a current source where will the resistance be connected?

A. In series with the current source
B. In parallel with the current source
C. The resistance is removed from the circuit
D. Resistance is multiplied by 10 and connected in parallel with the source

Answer: B

When we perform source transformation on a circuit, we transform a voltage source connected in series with a resistor to a current source connected in parallel to it. This is due to the relation we get by Ohm’s law, which is

V = IR.

147. A current source connected in parallel with a resistor can be converted to a?

A. Current source in series with a resistor
B. Voltage source in series with a resistor
C. Voltage source in parallel with a resistor
D. Cannot be modified

Answer: B

A current source connected in parallel can be converted to a voltage source connected in series using the relation obtained from Ohm’s law, that is V = IR.

This equation shows that a current source connected in parallel has the same impact as a voltage source connected in series.

148. A source transformation is _________

A. Unilateral
B. Bilateral
C. Unique
D. Cannot be determined

Answer: B

A source transformation is bilateral because a voltage source can be converted to a current source and vice − versa.

149. In source transformation________

A. Voltage source remains the same
B. Current sources remain the same
C. Both voltage and current source remain the same
D. Resistances remain the same

Answer: D

In source transformation, the value of the voltage and current sources change when changed from voltage to current source and current to voltage source but the value of the resistance remains the same.

150. If there are 3 10V sources connected in parallel then on source transformation __________

A. The effect of all the sources is considered
B. The effect of only one source is considered
C. The effect of none of the sources is considered
D. The effect of only 2 sources is considered.

Answer: B

When voltages are connected in parallel, the effect of only one source is considered because the effect of the voltage remains the same when connected in parallel.

151. The value of the three resistances when connected in star connection is _________

A. 2.32ohm,1.22ohm, 4.54ohm
B. 3.55ohm, 4.33ohm, 5.67ohm
C. 2.78ohm, 1.67ohm, 0.83ohm
D. 4.53ohm, 6.66ohm, 1.23ohm

Answer: C

Following the delta to star conversion:

R1 = 10 × 5/(10 + 5 + 3) = 2.78 ohm
R2 = 10 × 3/(10 + 5 + 3) = 1.67 ohm
R3 = 5 × 3/(10 + 5 + 3) = 0.83 ohm.

152. Which, among the following is the right expression for converting from delta to star?

R1 = Ra × Rb/(Ra + Rb + RC., R2 = Rb × Rc/(Ra + Rb + RC., R3 = Rc × Ra/(Ra + Rb + RC)
R1 = Ra/(Ra + Rb + RC., R2 = Rb/(Ra + Rb + RC., Rc = /(Ra + Rb + RC)
R1 = Ra × Rb × Rc/(Ra + Rb + RC., R2 = Ra × Rb/(Ra + Rb + RC., R3 = Ra/(Ra + Rb + RC)
R1 = Ra × Rb × Rc/(Ra + Rb + RC., R2 = Ra × Rb × Rc/(Ra + Rb + RC., R3 = Ra × Rb × Rc/(Ra + Rb + RC)

Answer: A

After converting to a star, each star-connected resistance is equal to the ratio of the product of the resistances it is connected to and the total sum of the resistances. Hence

R1 = Ra × Rb/(Ra + Rb + RC)
R2 = Rb × Rc/(Ra + Rb + RC)
R3 = Rc × Ra/(Ra + Rb + RC)

153. Find the equivalent star network using Delta-star Transformation.

A. 2.3ohm, 2.3ohm, 2.3ohm
B. 1.2ohm, 1.2ohm, 1.2ohm
C. 3.3ohm, 3.3ohm, 3.3ohm
D. 4.5ohm, 4.5ohm, 4.5ohm

Answer: B

The 6 ohm and 9-ohm resistances are connected in parallel. Their equivalent resistances are:

6 × 9/(9 + 6) = 3.6 ohm.

The 3 3.6 ohm resistors are connected in delta.

Converting to star:

R1 = R2 = R3 = 3.6 × 3.6/(3.6 + 3.6 + 3.6) = 1.2 ohm.

154. Star connection is also known as__________

A. Y − connection
B. Mesh connection
C. Either Y − connection or mesh connection
D. Neither Y − connection nor mesh connection

Answer: A

The star connection is also known as the Y-connection because its formation is like the letter Y.

155. Rab is the resistance between the terminals A and B, Rbc between B and C, and Rca between C and A. These 3 resistors are connected in a delta connection. After transforming to star, the resistance at A will be?

A. Rab × Rac/(Rab + Rbc + RcA)
B. Rab/(Rab + Rbc + RcA)
C. Rbc × Rac/(Rab + Rbc + RcA)
D. Rac/(Rab + Rbc + RcA)

Answer: A

When converting from delta to star, the resistances in star connection are equal to the product of the resistances it is connected to, divided by the total sum of the resistance.

Hence

Rab × Rac/(Rab + Rbc + RcA)

156. Rab is the resistance between the terminals A and B, Rbc between B and C, and Rca between C and A. These 3 resistors are connected in a delta connection. After transforming to star, the resistance at B will be?

A. Rac/(Rab + Rbc + RcA)
B. Rab/(Rab + Rbc + RcA)
C. Rbc × Rab/(Rab + Rbc + RcA)
D. Rab/(Rab + Rbc + RcA)

Answer: C

When converting from delta to star, the resistances in star connection are equal to the product of the resistances it is connected to, divided by the total sum of the resistance.

Hence Rab × Rbc/(Rab + Rbc + RcA)

157. Rab is the resistance between the terminals A and B, Rbc between B and C, and Rca between C and A. These 3 resistors are connected in a delta connection. After transforming to star, the resistance at C will be?

A. Rac/(Rab + Rbc + RcA)
B. Rab/(Rab + Rbc + RcA)
C. Rbc × Rac/(Rab + Rbc + RcA)
D. Rab/(Rab + Rbc + RcA)

Answer: C

When converting from delta to star, the resistances in star connection are equal to the product of the resistances it is connected to, divided by the total sum of the resistance.

Hence Rac × Rbc/(Rab + Rbc + RcA)

158. Find the current in the circuit using Delta star Transformation.

A. 0.54A
B. 0.65A
C. 0.67A
D. 0.87A

Answer: A

The 35-ohm resistors are connected in delta.

Changing it to star:

R1 = R2 = R3 = 1.67 ohm.

One of the 1.67- ohm resistors is connected in series with the 2-ohm resistor and another 1.67-ohm resistor is connected in series to the 3 − ohm resistor.

The resulting network has a 1.67 − ohm resistor connected in series with the parallel connection of the 3.67 and 4.67 resistors.

The equivalent resistance is 3.725A.

I = 2/3.725 = 0.54A.

159. If a 6 ohm, 2ohm, and 4ohm resistor is connected in delta, find the equivalent star connection.

A. 1ohm, 2ohm, 3ohm
B. 2ohm, 4ohm, 7ohm
C. 5ohm, 4ohm, 2ohm
D. 1ohm, 2ohm, 2/3ohm

Answer: D

Using the delta to star conversion formula:

R1 = 2 × 6/(2 + 6 + 4)
R2 = 2 × 4/(2 + 6 + 4)
R3 = 4 × 6/(2 + 6 + 4).

160. If a 4ohm, 3ohm, and 2ohm resistor is connected in delta, find the equivalent star connection.

A. 8/9ohm, 4/3ohm, 2/3ohm
B. 8/9ohm, 4/3ohm, 7/3ohm
C. 7/9ohm, 4/3ohm, 2/3ohm
D. 8/9ohm, 5/3ohm, 2/3ohm

Answer: A

Using the delta − star conversion formula:

R1 = 4 × 3/(2 + 3 + 4)
R2 = 2 × 3/(2 + 3 + 4)
R3 = 2 × 4/(2 + 3 + 4)

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