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Right CHOICE is (a) 50 W

Explanation: The THEVENIN EQUIVALENT of the circuit is as shown below.

I = 10 A, RTH = 2 Ω

∴ Pmax = (\(\frac{10}{2}\))^2 × 2

= 5×5×2 = 50 W.

Right CHOICE is (b) 7 kΩ

For EXPLANATION: LET VAB = 1 V

5 VAB = 5

Or, 1 = 1 × I1 or, I1 = 1

Also, 1 = -5 + 1(I – I1)

∴ I = 7

Hence, R = 0.2 kΩ.

The correct choice is (b) 5 V and 6 Ω

Best explanation: We, Thevenized the left side of xx’ and source transformed right side of yy’.

VXX’ = Vth = \(\displaystyle\frac{\frac{4}{8} + \frac{8}{24}}{\frac{1}{8} + \frac{1}{24}}\) = 5V

∴ RTH = 8 || (16 + 8)

= \(\frac{8×24}{8+24}\) = 6 Ω.

Right answer is (C) 1.5

For EXPLANATION I WOULD SAY: 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.

Correct OPTION is (b) 1.5

For 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.

The correct CHOICE is (C) 3

The EXPLANATION is: 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.

Correct ANSWER is (c) single current source I’ in parallel with G’

The best explanation: Millman’s THEOREM STATES that if there are current sources I1,I2,…… In with internal CONDUCTANCES G1,G2,…..Gn, respectively, are in series, then these sources are replaced by single current source I’ in parallel with G’.

Correct CHOICE is (c) R’=1/((G1+G2+⋯Gn))

For EXPLANATION: Let the equivalent resistance is R’. The value of equivalent resistance is R’=1/((G1+G2+⋯Gn)).

Right ANSWER is (b) single voltage SOURCE V’ in SERIES with R’

To explain I would say: Millman’s Theorem states that if there are voltage SOURCES V1, V2,…… Vn with internal resistances R1, R2,…..Rn, respectively, are in parallel, then these sources are replaced by single voltage source V’ in series with R’.

Correct choice is (a) \(\Sigma^{B}_{k=1}\)V1ki2k = 0, \(\Sigma^{b}_{k=1}\)V2ki1k = 0

To explain: If two networks have same graph with different TYPE of elements between corresponding NODES, then \(\Sigma^{b}_{k=1}\)V1ki2k = 0, \(\Sigma^{b}_{k=1}\)V2ki1k = 0.

The CORRECT ANSWER is (a) satisfies

Explanation: i1=i2=2A, i3=2A. V1=-2V, V2=-8V, V3=10V. ALGEBRAIC SUM =

The correct answer is (a) True

For EXPLANATION: i1=i2=4A, i3=4A. V1=-20V, V2=0V, V3=20V. Algebraic SUM =

Correct answer is (c) equal

For EXPLANATION I would say: For TELLEGAN’s Theorem to satisfy, algebraic sum of the POWER delivered by the source equal to power absorbed by all elements. All branch CURRENTS and voltages in that network must satisfy Kirchhoff’s laws.

Right answer is (a) KVL

Best EXPLANATION: In an electrical CIRCUIT itself there are pairs of TERMS which can be interchanged to get new CIRCUITS. The dual pair of KCL is KVL. And the dual pair of KVL is KCL.

The CORRECT option is (d) all of the mentioned

For explanation I would SAY: Tellegen’s Theorem is valid for any lumped network. So, Tellegan’s theorem is valid for linear or non-linear networks, passive or active networks and time VARIANT or time invariant networks.

The CORRECT option is (b) CURRENT source

Easiest explanation: The DUAL pair of voltage source is current source. And the dual pair of current source is voltage source.

The CORRECT option is (c) CONDUCTANCE

To EXPLAIN: The dual pair of resistance is conductance. And the dual pair of conductance is resistance.

Right answer is (d) INDUCTANCE

To EXPLAIN I would SAY: The dual pair of inductance is capacitance. And the dual pair of capacitance is inductance. In an electrical CIRCUIT itself, there are pairs of terms which can be interchanged to get NEW circuits.

The CORRECT answer is (C) 98.6

Easy EXPLANATION: ZT = RS – J5+ RL = 2-j5+20 = 22.56∠-12.8⁰Ω. I=VS/ZT = -50∠0⁰/22.56∠-12.8⁰ = 2.22∠-12.8⁰A. P = I^2R = (2.22)^2×20 = 98.6W.

Right option is (a) voltage

To explain I WOULD say: In an electrical circuit itself there are pairs of terms that can be interchanged to get new CIRCUITS. The DUAL PAIR of current is voltage. And the dual pair of voltage is current.

The correct option is (b) 2

The BEST explanation: As RL is fixed, the MAXIMUM power TRANSFER theorem does not APPLY. Maximum current FLOWS in the circuit when RS is minimum. So RS = 2Ω.

Correct answer is (c) 41.33

Best explanation: The term power is defined as the PRODUCT of the square of CURRENT and the IMPEDANCE. So the maximum power delivered by the source in the CIRCUIT is P = I^2RxZ = 1.66^2×15 = 41.33W.

The correct answer is (B) 1.66∠0⁰

The BEST explanation: The load CURRENT is the RATIO of voltage to the impedance. So the load current is I = (50∠0^o)/(15+j20+15-j20) = 1.66∠0^o A.

The correct CHOICE is (a) 15-j20

To EXPLAIN: The maximum power transfer occurs when the load IMPEDANCE is equal to the complex conjugate of SOURCE impedance ZS. ZL = ZS^* = (15-j20)Ω.

Correct answer is (b) ZL = ZS*

Best explanation: The MAXIMUM power is TRANSFERRED when the load resistance is EQUAL to the source resistance. The condition for maximum power TRANSFER is ZL= ZS*.

The correct choice is (c) complex conjugate of

To explain I would say: The MAXIMUM power transfer theorem can be APPLIED to complex IMPEDANCE circuits. If source impedance is complex, the maximum power transfer occurs when the LOAD impedance is complex conjugate of the source impedance.

The correct choice is (c) EQUAL to

Easy explanation: The maximum POWER is delivered from a SOURCE to its load when the load RESISTANCE is equal to the source resistance. The maximum power TRANSFER theorem can be applied to both dc and ac circuits.

The CORRECT CHOICE is (a) 0.6

The explanation: The current FLOWING in the ammeter having 1Ω internal RESISTANCE in series with the 3Ω resistor shown in the circuit is 0.6 A.

Current through 3Ω resistor = 0.67/(7 + (4||6||2)) = 0.08A. Ammeter READING = 0.67 – 0.08 = 0.59 ≅ 0.6A.

Correct choice is (c) 0.67

Easy explanation: TOTAL current = 10 / (4 + (6||2||3) = 2A. Current through 3Ω RESISTOR= 2 x (6||2)/(3 + (6||2)) = 0.67A.

The CORRECT ANSWER is (c) 0.3

To explain: NEW TOTAL current = 0.33/(7+4||2||3)=0.04A. Now reading of ammeter = 0.33-0.04=0.29A ≅ 0.3A.

Correct option is (C) 0.33

Best EXPLANATION: TOTAL CURRENT in the circuit = 10/(4+3||2||6)=2A. Current through 6Ω RESISTOR = 2×(3||2)/(6+3||2)=0.33A.

The CORRECT choice is (c) 0.55

To EXPLAIN I would say: Total resistance in the circuit = 4+6||3Ω. The total current drawn by the circuit = 10/(4+6||3)=1.67A. Current through 6Ω resistor = 0.55A.

Right option is (c) 0.93

Best explanation: CURRENT through 3Ω resistor = 1.11A. So voltage DROP across 1Ω resistor = 1.11×1 = 1.11V. Now the circuit can be modified as

Now current through 3Ω resistor = 0.17A. This current is OPPOSITE to the current CALCULATED before. So ammeter reading = (1.11-0.17) = 0.94A.

The correct ANSWER is (a) 1

To EXPLAIN I would say: Total resistance in the CIRCUIT = 2+[3||(2+2||2)] = 3.5Ω. The total CURRENT drawn by the circuit =10/(4+6||3) = 1.67A. Current through 3Ω RESISTOR = 1.11A ≅1A.

The correct OPTION is (C) voltage to current

For explanation: While considering Reciprocity THEOREM, we consider RATIO of response to excitation as ratio of voltage to current or current to voltage.

The correct option is (c) Linear BILATERAL

The explanation: RECIPROCITY THEOREM is applied for linear bilateral NETWORKS, not for linear or for linear bilateral or for LUMPED networks.

Correct ANSWER is (c) 2

The explanation: The circuit after SOURCE is replaced is

Total current from the source (It)=12/(3+(6│|2))=2.67A. Current through 2Ω resistor=2.67× 6/(6+2)=2A.

Right choice is (b) 2

The explanation is: The 6Ω resistor is PARALLEL to 3Ω resistor and the RESULTANT is in series with 2Ω resistor. Total CURRENT from source = 12/(2+(6│|3))=3A. Current through 3Ω resistor = 3 × 6/(6+3)=2A.

Correct ANSWER is (c) 1.43

For EXPLANATION: The CIRCUIT after the source is replaced is

Total resistance = 3.23Ω. The CURRENT drawn by the circuit(It)=20/3.23=6.19A. The current in branch a-b is 1.43A.