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The CORRECT choice is (b) 2 A

To explain: Equivalent Resistance, REQ = 20 + [60 || 30]

= 20 + \(\frac{60 × 30}{60+30}\)

= 20 + 20 = 40 Ω

Total CURRENT from the source, I = \(\frac{120}{40}\) = 3A

Now, by USING current division rule, I3Ω = \(\frac{3 × 60}{30+60}\) = 2 A.

Correct choice is (a) The circuit follows Reciprocity Theorem

The explanation: Let us CONSIDER this circuit,

Equivalent Resistance REQ = 20 + [30 || (20 + (20||20))]

= 20 + [30 || (20 + \(\FRAC{20×20}{20+20}\))]

= 20 + [30 || (20+10)]

= 20 + [30 || 30]

= 20 + \(\frac{30 × 30}{30+30}\)

= 20 + 15 = 35 Ω

The current drawn by the circuit = \(\frac{200}{35}\) = 5.71 A

Now, by using current division rule, we get, I1 = 1.43 A

Again, let us consider this circuit,

Equivalent Resistance, REQ = [[((30 || 20) + 20) || 20] + 20]

= \(\BIG[\Big[\left(\left(\frac{30 × 20}{30+20}\right) + 20\right) || 20\Big] + 20\Big]\)

= [[(12 + 20) || 20] + 20]

= [[32 || 20] + 20]

= \(\Big[\left(\frac{32 × 20}{32+20}\right) + 20\Big]\)

= [12.31 + 20] = 32.31 Ω

The current drawn by the circuit = \(\frac{200}{32.31}\) = 6.19 A

Now, by using current division rule, we get, I2 = 1.43 A.

Since I1 = I2, the circuit follows Reciprocity Theorem.

Correct OPTION is (d) 1.43 A

To explain I would say: Equivalent RESISTANCE, REQ = [[((30 || 20) + 20) || 20] + 20]

= \(\Big[\Big[\left(\left(\frac{30 × 20}{30+20}\right) + 20\right) || 20\Big] + 20\Big]\)

= [[(12 + 20) || 20] + 20]

= [[32 || 20] + 20]

= \(\Big[\left(\frac{32 × 20}{32+20}\right) + 20\Big]\)

= [12.31 + 20] = 32.31 Ω

The current drawn by the CIRCUIT = \(\frac{200}{32.31}\) = 6.19 A

Now, by using current division rule, we get, I2Ω = 1.43 A.

Right ANSWER is (c) 1.43 A

Explanation: EQUIVALENT Resistance REQ = 20 + [30 || (20 + (20||20))]

= 20 + [30 || (20 + \(\frac{20×20}{20+20}\))]

= 20 + [30 || (20+10)]

= 20 + [30 || 30]

= 20 + \(\frac{30 × 30}{30+30}\)

= 20 + 15 = 35 Ω

The current DRAWN by the circuit = \(\frac{200}{35}\) = 5.71 A

Now, by USING current division rule, we get, I2Ω = 1.43 A.

Right CHOICE is (a) 1.5 A

Explanation: Equivalent RESISTANCE, REQ = [(12 || 6) + 2 + 4] = 10 Ω

IS = \(\FRAC{45}{10}\) = 4.5 A

Now, by using Current division rule, we get, I = \(\frac{4.5 × 6}{12+6} = \frac{27}{18}\) = 1.5 A.

Correct option is (a) The circuit follows Reciprocity Theorem

Easy explanation: Let us consider this circuit,

Rth = [(2+4) || 6] + 12 = 15 Ω

IS = \(\frac{45}{15}\) = 3 A

Now, by current division rule, we GET, I1 = \(\frac{3 × 6}{12} = \frac{18}{12}\) = 1.5 A.

Again, let us consider this circuit,

Equivalent resistance,REQ = [(12 || 6) + 2 + 4] = 10 Ω

IS = \(\frac{45}{10}\) = 4.5 A

Now, by using Current division rule, we get, I2 = \(\frac{4.5 × 6}{12+6} = \frac{27}{18}\) = 1.5 A.

Since I1 = I2, the circuit follows Reciprocity Theorem.

The correct ANSWER is (a) Single-source networks

Explanation: ACCORDING to Reciprocity THEOREM, the voltage source and the RESULTING current source MAY be interchanged without a change in current. Therefore the theorem is applicable only to single-source networks. It therefore cannot be employed in multi-source networks.

The CORRECT choice is (c) 1.5 A

For explanation I would SAY: RTH = [(2+4) || 6] + 12 = 15 Ω

IS = \(\frac{45}{15}\) = 3 A

Now, by current division rule, we GET, I = \(\frac{3 × 6}{12} = \frac{18}{12}\) = 1.5 A.

Correct option is (B) Linear TIME Invariant circuits

To elaborate: A reciprocal network comprises of linear time-invariant bilateral elements. It is applicable to resistors, capacitors, INDUCTORS (INCLUDING coupled inductors) and transformers. However, both dependent and INDEPENDENT sources ate not permissible.

Correct choice is (a) 50 W

For explanation: Using Superposition Theorem, we GET, I = 10 A and R = 2 Ω

∴ Pmax = (\(\frac{10}{2}\))^2 × 2

= 5 × 5 × 2 = 50 W.

Right option is (b) \(\FRAC{8}{3}\) Ω

To explain I would SAY: USING Superposition Theorem, in the given circuit, \(\frac{V}{4} + \frac{V-2I}{2}\) = I

Or, \(\frac{3V-4I}{4}\) = I

Or, 3V = 8I

∴ \(\frac{V}{I} = \frac{8}{3}\) Ω.

The correct choice is (c) V = \(\frac{2}{7} V_S – \frac{4}{7} I_S\)

Explanation: By SUPERPOSITION, the current I is given by I = \(\frac{V_S}{5} − \frac{2}{5} × I_S − \frac{3}{5}× 3i\)

This can be solved for I to OBTAIN, I = \(\frac{V_S}{14} – \frac{I_S}{7}\)

Now, by Superposition Theorem, V = \(\frac{V_S}{5} − \frac{2}{5} × I_S − \frac{3}{5}× 3i\)

Or, V = \(\frac{V_S}{5} − \frac{2}{5} × I_S − \frac{3}{5}× 3(\frac{V_S}{14} – \frac{I_S}{7})\)

∴ V = \(\frac{2}{7} V_S – \frac{4}{7} I_S\).

Correct option is (a) 24 V

Best EXPLANATION: Using Superposition Theorem, we get,

V∆ = −0.4V∆ × 10 + 5 × 10

Solving for V∆, we get, V∆ = \(\frac{5 × 10}{1+0.4×10}\) = 10 V

Again by using Superposition Theorem, we get, I∆ = \(\frac{10}{5+20} – 0.4V_∆\frac{20}{5+20}\)

Or, I∆ = \(\frac{10}{25} – 0.4V_∆\frac{20}{25} = -\frac{70}{25}\) A

Thus, VO is GIVEN by, VO = 10 − 5I∆

∴ VO = 24 V.

Correct ANSWER is (c) −30 + 10IO

Easy explanation: Using Superposition Theorem the VOLTAGE VX is GIVEN as,

VX = (3-IO) (2||40) + 5VX \(\frac{2}{40+2}\)

Or, VX = \(\frac{80}{42}(3 – I_O) + \frac{10}{42}\)VX

Or, VX = \(\frac{\frac{80}{42}}{1-\frac{10}{42}}\)(3 – IO) = 2.5(3 – IO)

∴ VO = VX – 5VX = −30 + 10IO.

Correct answer is (b) 30 V

For explanation: Using Superposition Theorem, the current IB is given by,

IB = \(\frac{70}{4||20+2||10} × \frac{20}{4+20} + \frac{50}{10+4||20||2} × \frac{20||2}{4+20||2} – \frac{2I_B}{20||2+4||10} × \frac{10}{4+10}\)

Or, IB = \(\frac{35}{3} + \frac{25}{18} – \frac{11}{36}\) IB

Now, Solving for IB, we get, IB = \(\frac{\frac{35}{3}+\frac{25}{18}}{1+\frac{11}{36}}\) = 10 A

Now, VO = 70 − 4ib

∴ VO = 30 V.

The correct answer is (d) 2.5(3 – IO)

BEST explanation: Using SUPERPOSITION Theorem the voltage VX is given as,

VX = (3-IO) (2||40) + 5VX \(\frac{2}{40+2}\)

Or, VX = \(\frac{80}{42}(3 – I_O) + \frac{10}{42}V_X\)

Or, VX = \(\frac{\frac{80}{42}}{1-\frac{10}{42}}(3 – I_O)\) = 2.5(3 – IO).

Right choice is (b) 20

To elaborate: Using Superposition theorem, we get,

\(\frac{V_P-100}{10} + \frac{V_P}{10}\) + 2 = 0

Or, 2VP – 100 +20= 0

∴ VP = 80/2 = 40V

∴ R = 20Ω.

The correct choice is (a) 11.148 V

Explanation: The datum NODE is the LOWER branch.

 By superposition, the current i is given by, i = 2 \(\frac{7}{15+5+7} + \frac{3}{15+5+7} + 4I \frac{7+15}{15+5+7}\)

Or, I = \(\frac{17}{27} + \frac{88}{27I}\)

∴ The solution for i yields, I = \(\frac{\frac{17}{27}}{1 – \frac{88}{27}} = -\frac{17}{61}\) A

∴ V1 = V2 – (4I – i) 15 = 11.148 V.

The correct OPTION is (c) -1.393 V

The best explanation: The DATUM node is the lower branch.

By superposition, the CURRENT i is given by, i = 2 \(\frac{7}{15+5+7} + \frac{3}{15+5+7} + 4I \frac{7+15}{15+5+7}\)

Or, I = \(\frac{17}{27} + \frac{88}{27I}\)

∴ The solution for i yields, I = \(\frac{\frac{17}{27}}{1 – \frac{88}{27}} = -\frac{17}{61}\) A

∴ V2 = 5I = -1.393 V.

Right answer is (b) \(\frac{5}{4}\) A

Explanation: Using SUPERPOSITION Theorem, we can write

I = \(\frac{24}{3+2} – \frac{72}{3+2} – \frac{3I}{3+2} \)

Or, I = 2 – \(\frac{3I}{3+2}\)

Or, I = 2 – \(\frac{3I}{5}\)

∴ I = \(\frac{5}{4}\) A.

Correct option is (a) Power CALCULATION

Explanation: The Superposition Theorem is not applicable for Power calculation because for power, the calculations involve EITHER the product of voltage and CURRENT or the square of current or the square of the voltage thus making them non-linear OPERATIONS. HENCE they cannot be calculated using Superposition Theorem.

The correct option is (d) 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

Best explanation: 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.

Correct option is (c) 1 A

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

I1 = \(\FRAC{3}{2+1}\) = 1 A(B -> A)

The net current in 2 Ω resistance is I = 1 – I1

= 2 – 1 = 1 A(A -> B).

The correct option is (d) 12.12 Ω

The best explanation: VX = 3\(\frac{V_X}{6}\) + 4

Or, VX = 8 V = VOC

If terminal is short-circuited, VX = 0.

ISC = \(\frac{4}{3+3}\) = 0.66 A

∴ RN = \(\frac{V_{OC}}{I_{SC}}\)= \(\frac{8}{0.66}\) = 12.12 Ω.

The correct option is (d) 6 Ω and 1.167 A

For explanation: 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

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

= \(\frac{0.167+1}{0.04167+0.125}\) = 7 V

∴ RN = (8 + 16) || 8

= \(\frac{24×8}{24+8}\) = 6 Ω

∴ IN = \(\frac{V_N}{R_N} = \frac{7}{6}\) = 1.167 A.

Correct option is (b) 9.26 W

Easy explanation: Let us remove the 1 Ω resistor and short x-y.

At Node 1, assuming node potential to be V, \(\frac{V-10}{5}\) + ISC = 5

But ISC = \(\frac{V}{2}\)

∴\(\frac{V-10}{5} + \frac{V}{2}\) = 5

Or, 0.7 V = 7

That is, V= 10 V

∴ ISC = \(\frac{V}{2}\) = 5 A

To find Rint, all constant sources are deactivated. Rint = \(\frac{(5+2)×2}{5+2+2} = \frac{14}{9}\) = 1.56 Ω

Rint = 1.56 Ω; ISC = IN = 5A

Here, I = IN \(\frac{R_{INT}}{R_{int}+1} = 5 × \frac{1.56}{1.56+1}\) = 3.04 A

∴ Power LOSS = (3.04)^2 × 1 = 9.26 W.

Correct option is (b) 6 Ω and 0.833 A

To ELABORATE: We, draw the NORTON equivalent of the left side of xx’ and source TRANSFORMED right side of YY’.

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

∴ RN = 8 || (16 + 8)

= \(\frac{8×24}{8+24}\) = 6 Ω

∴ \(I_N = \frac{V_N}{R_N} = \frac{5}{6}\) = 0.833 A.

Right ANSWER is (b) 30 kΩ

The best I can EXPLAIN: When RL = 10 kΩ and VAB = 4 V

Current in the circuit \(\frac{V_{AB}}{R_L} = \frac{4}{10}\) = 0.4 mA

Norton voltage is given by VN = I (RN + RL)

= 0.4(RN + 10)

= 0.4RN + 4

Similarly, for RL = 2 kΩ and VAB = 1 V

So, I = \(\frac{1}{2}\) = 0.5 mA

VN = 0.5(RN + 2)

= 0.5 RN + 1

∴ 0.1RN = 3

Or, RN = 30 kΩ.

The correct answer is (a) 1

The EXPLANATION is: To calculate the Norton resistance, all the CURRENT sources GET open-circuited and VOLTAGE sources get short-circuited.

∴ RN = (\(\FRAC{1}{s}\)+ 1) || (1+s)

= \(\frac{\left(\frac{1}{s} + 1\right)×(1+s)}{\left(\frac{1}{s} + 1\right)+(1+s)}\)

= \(\frac{\frac{1}{s}+1+1+s}{\frac{1}{s}+1+1+s}\) = 1

So, RN = 1.

Correct choice is (a) 5 Ω

Easy explanation: For finding VN,

VN = \(\frac{4 × 10}{10+10}\) = 2V

For finding RN,

RN = 10 || 10

= \(\frac{10×10}{10+10}\) = 5 Ω.

The correct CHOICE is (d) 2 Ω

The explanation is: \(R_N = \frac{V_{OC}}{I_{SC}}\)

VN = VOC

Applying KCL at node A, \(\frac{2I-V_N}{1} + 2 = I + \frac{V_N}{2}\)

Or, I = \(\frac{V_N}{1}\)

PUTTING, 2VN – VN + 2 = VN + \(\frac{V_N}{2}\)

Or, VN = 4 V.

∴ RN = 4/2 = 2Ω.

The correct option is (a) 100 Ω

The EXPLANATION: IX = 1 A, VX = Vtest

Vtest = 100(1-2IX) + 300(1-2IX – 0.01VS) + 800

Or, Vtest = 1200 – 800IX – 3Vtest

Or, 4Vtest = 1200 – 800 = 400

Or, Vtest = 100V

∴ RN = \(\frac{V_{TEST}}{1}\) = 100 Ω.

Correct option is (C) 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

To EXPLAIN I would say: While computing the Thevenin equivalent voltage consisting of both dependent and independent sources, we first find the equivalent voltage called the Thevenin voltage by opening the TWO terminals. Then while computing the Thevenin equivalent RESISTANCE, 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.

The correct option is (a) 1

Explanation: To calculate the Thevenin resistance, all the CURRENT sources GET open-circuited and voltage source short-circuited.

∴ RTH = (\(\frac{1}{s}\)+ 1) || (1+s)

= \(\frac{\left(\frac{1}{s} + 1\right)×(1+s)}{\left(\frac{1}{s} + 1\right)+(1+s)}\)

= \(\frac{\frac{1}{s}+1+1+s}{\frac{1}{s}+1+1+s}\) = 1

So, RTH = 1.

Right choice is (a) 2 V and 5 Ω

Easiest explanation: For finding VTH,

VTH = \(\frac{4 ×10}{10+10}\) = 2V

For finding RTH,

RTH = 10 || 10

= \(\frac{10×10}{10+10}\) = 5 Ω.

Correct answer is (B) 20

Easy EXPLANATION: \(\frac{V_P-100}{10} + \frac{V_P}{10}\) + 2 = 0

Or, 2VP – 100 + 20 = 0

∴ VP = 80/2 = 40V

∴ R = 20Ω.

The correct option is (d) 4 V and 2 Ω

The explanation is: \(R_{TH} = \frac{V_{OC}}{I_{SC}}\)

VTH = VOC

Applying KCL at NODE A, \(\frac{2I-V_{TH}}{1} + 2 = I + \frac{V_{TH}}{2}\)

Or, I = \(\frac{V_{TH}}{1}\)

PUTTING, 2VTH – VTH + 2 = VTH + \(\frac{V_{TH}}{2}\)

Or, VTH = 4 V.

∴ RTH = 4/2 = 2Ω.

Correct choice is (C) 30 V

To explain I would say: By applying KCL, \(\frac{V_P-E}{6} + \frac{V_P}{6}\) – 1 = 0

Or, 2 VP – E = 6

Where, (\(\frac{-V_P+E}{6}\)) = 2

∴ E – VP = 12

Or, VP = 18 V

∴ E = 30V.

Right choice is (b) 30 kΩ and 16 V

To ELABORATE: When RL = 10 kΩ and VAB = 4 V

Current in the circuit I = \(\frac{V_{AB}}{R_L} = \frac{4}{10}\) = 0.4 mA

Thevenin voltage is given by VTH = I (RTH + RL)

= 0.4(RTH + 10)

= 0.4RTH + 4

Similarly, for RL = 2 kΩ and VAB = 1 V

So, I = \(\frac{1}{2}\) = 0.5 mA

VTH = 0.5(RTH + 2)

= 0.5 RTH + 1

∴ 0.1RTH = 3

Or, RTH = 30 kΩ

And VTH = 12 + 4 = 16 V.

The correct ANSWER is (a) 100 Ω

Explanation: IX = 1 A, VX = Vtest

Vtest = 100(1-2IX) + 300(1-2IX – 0.01VS) + 800

Or, Vtest = 1200 – 800IX – 3Vtest

Or, 4Vtest = 1200 – 800 = 400

Or, Vtest = 100V

∴ RTH = \(\frac{V_{test}}{1}\) = 100 Ω.

The correct CHOICE is (d) VOC = 0 V, RTH = 90 Ω

To elaborate: By writing loop equations for the circuit, we GET,

VS = VX, IS = IX

VS = 600(I1 – I2) + 300(I1 – I2) + 900 I1

= (600+300+900) I1 – 600I2 – 300I3

= 1800I1 – 600I2 – 300I3

I1 = IS, I2 = 0.3 VS

I3 = 3IS + 0.2VS

VS = 1800IS – 600(0.01VS) – 300(3IS + 0.01VS)

= 1800IS – 6VS – 900IS – 3VS

10VS = 900IS

For Thevenin equivalent, VS = RTH IS + VOC

So, Thevenin voltage VOC = 0

Resistance RTH = 90Ω.

The correct choice is (d) 0.5 V

Easiest explanation: Current through 1 Ω = \(\frac{5}{2}\) – I1

Using SOURCE transformation to 5 V SOURCES, VOC = 1 × I1

VOC = -5 VOC + (\(\frac{5}{2}\) – I1) × 1

Eliminating I1, we get, VOC = 0.5 V.