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The correct CHOICE is (c) 113.04 μH

The explanation: INDUCTANCE of the COIL, L = \(\frac{μ_0 n^2 A}{l}\)

= \(\frac{4π X 10^{-7}X 300 X 300 X 300 X 10^{-6}}{300 X 10^{-3}}\)

= 113.04 μH.

The correct option is (d) 1.57

For explanation I would say: From the voltage EQUATION, we can get Vm = 150 V and Im = 150 / 20 = 7 A

RMS value of the CURRENT, Irms = Im / 2 = 7/2 = 3.5 A

Average value of the current, Iavg = Im / π = 2.228 A

Form factor = Irms / Iavg = 3.5 / 2.228 = 1.57.

The correct answer is (c) 200 V

The best explanation: GIVEN that, potential difference between ANODE and CATHODE (earthed) is MEASURED as 150 V and potential of EARTH is – 50 V.

So, actual voltage on anode, V = 150 – (- 50)

= 150 + 50

= 200 V.

The correct option is (b) 141.4 A

For explanation I would say: We know that for SINUSOIDAL alternating electric CURRENT the peak factor or amplitude factor can be expressed the ratio of maximum or peak value and RMS value of alternating current.

So the peak value = rms value of alternating electric current × peak factor of alternating electric current = 100 × 1.414 = 141.4 A.

Right ANSWER is (b) 76.68 W

The best I can explain: I2 = \(\frac{V_2}{6}\), I1 = \(\frac{I_2}{5} = \frac{V_2}{30}\)

V1 = 5V2

50 = 400(I1 – 0.04V2) + V1

Or, V2= 21.45 V

∴ PL = \(\frac{V_2^2}{6} \)

= \(\frac{21.45^2}{6} \)

= \(\frac{460.1025}{6}\) = 76.68 W.

Correct answer is (c) 1.11

Best EXPLANATION: We know that for alternating electric CURRENT form FACTOR is defined as the RATIO of rms value and average value of alternating current.

Now the rms value of alternating electric current = 0.07 × MAXIMUM value of alternating current.

Average value of alternating electric current = 0.637 × maximum value of alternating current.

∴ Form factor = \(\frac{0.707}{0.637}\) = 1.11.

The correct answer is (a) 1 W

The best EXPLANATION: 3 = 2I + I^2

∴ I = 1 A; VNL = 1V

∴ Power DISSIPATED in RNL = 1 × 1 = 1 W.

The CORRECT choice is (c) 350.38 Ω

Explanation: 5 = 200I – 50 × 2I

Or, I = \(\frac{5}{100}\) = 0.05 A

VOC = 100 × 3I + 200 × I = 25 V

V1 = \(\frac{\frac{5}{50}}{\frac{1}{50} + \frac{1}{200} + \frac{1}{100}} \)

= \(\frac{0.1}{0.02+0.005+0.01} \)

= 2.85 V

I = \(\frac{2.85}{100}\) = 0.0142 A = 14.2 mA

ISC = \(\frac{2.85}{100}\) + 3 × 0.0142 = 0.07135 A

∴ RTH = \(\frac{V_{OC}}{I_{SC}}= \frac{25}{0.07135}\) = 350.38 Ω.

Right option is (d) 66 W

Best EXPLANATION: Drop ACROSS V1Ω = 5 × 1 = 5V

Also, \(\frac{V-V_{1Ω}}{10} + \frac{V-20-V_{1Ω}}{2} + \frac{V-V_{OC}}{5}\) = 2

Or, 0.1 V – 0.1V1Ω + 0.5V – 10 – 0.5V1Ω + 0.2 – 0.2VOC = 2

Or, 0.8V – 0.6V1Ω = 12 + 0.2VOC

Or, 0.8 V – 0.2VOC = 12 +3=15 (Putting V1Ω = 5)

Again, \(\frac{V_{OC}-V}{5}\) + 2 = 5

Or, 0.2VOC – 0.2V = 3

Again, RTH = {(10||2) + 1} + 5

= (\(\frac{20}{12+1}\)) + 5 = 7.67 Ω

Following the theorem of maximum power transfer, R = RTH = 7.67 Ω

And PMAX = \(\frac{V_{OC}^2}{4R} = \frac{45^2}{4×7.67}\) = 66 W.

The correct answer is (C) It is a PERFECT transformer

The best EXPLANATION: We KNOW that, K = \(\frac{M}{\SQRT{L_1 L_2}}\)

= \(\frac{2}{\sqrt{2 X 2}}\) = 1

Hence, it is a perfect transformer.

Right choice is (b) \(\frac{4}{3}\) Ω

To elaborate: Given that copper and COATED metal strip have resistance of 2 ohms RESPECTIVELY. These two strips are connected in parallel.

Hence, the resistance of the COMPOSITE strip = \(\frac{2 X 4}{2 + 4}\)

= \(\frac{8}{6} = \frac{4}{3}\) Ω.

Right option is (C) 4 Ω

The explanation is: Maximum power is transferred to RL when the load resistance equals the THEVENIN resistance of the circuit.

RL = RTH = \(\frac{V_{OC}}{I_{SC}} \)

DUE to open-circuit, VOC = 100 V; ISC = I1 + I2

Applying KVL in LOWER loop, 100 – 8I1 = 0

Or, I1 = \(\frac{100}{8} = \frac{25}{2}\)

And VX = -4I1 = -4 × \(\frac{25}{2}\) = -50V

KVL in upper loop, 100 + VX – 4I2 = 0

I2 = \(\frac{100-50}{4} = \frac{25}{2}\)

Hence, ISC = I1 + I2 = \(\frac{25}{2} + \frac{25}{2}\) = 25

RTH = \(\frac{V_{OC}}{I_{SC}} = \frac{100}{25}\) = 4 Ω

RL = RTH = 4 Ω.

The correct answer is (b) 0.49 J

The explanation: L = \(\frac{N^2 μ_0A}{l}\)

= \(\frac{10^6.4π.10^{-7}.\frac{π}{4}.(100 X 10^{-4})}{1}\)

= \(\frac{π^2X 10^{-3}}{1}\)

Energy = 0.5 LI^2

= 0.49 J.

Right ANSWER is (a) 0.75 W

The EXPLANATION: I + 0.9 = 10 I

Or, I = 0.1 A

VOC = 3 × 10 I = 30 I

Or, VOC = 3 V

Now, ISC = 10 I = 1 A

Rth = 3/1 = 3 Ω

Vth = VOC = 3 V

RL = 3 Ω

Pmax = \(\FRAC{3^2}{4×3}\) = 0.75 W.

The correct answer is (c) 0.54 A

To explain: In the GIVEN figure, RX = RY = RZ = \(\frac{5×5}{5+5+5}\) = 1.67 Ω

Here, R = [(RX + 2) || (RY + 3)] + RX

= (3.67 || 4.67) + RZ

= \(\frac{3.67 × 4.67}{3.67 + 4.67}\) + 1.67

= \(\frac{17.1389}{8.34}\) + 1.67

= 3.725 Ω

∴ I = \(\frac{V}{R} = \frac{2}{3.725}\) = 0.54 A.

Correct choice is (a) R, L and C

Best EXPLANATION: The elements which show their BEHAVIOUR only when excited are called as basic circuit elements. Here resistance, INDUCTANCE and capacitance show their behaviour only when excited. Hence they are the basic elements of an electric circuit.

Right option is (d) 240 V

Explanation: In ORDER for 600 C charges to be delivered to 100 V source, the electric current must be anti-clockwise.

 \(i = \FRAC{dQ}{DT} = \frac{600}{60}\) = 10A

Applying KVL we get

V1 + 60 – 100 = 10 × 20 ⇒ V1 = 240 V.

Correct answer is (d) VOLTAGE and current source both

Easiest explanation: We know that ACTIVE elements are the ones that are used to drive the CIRCUIT. They ALSO cause the ELECTRIC current to flow through the circuit or the voltage drop across the element. Here only the voltage and current source are the ones satisfying the above conditions.

Correct answer is (C) 10 A

The best I can explain: AVERAGE DC electric CURRENT = 10 A

Average ac electric current = 0 A as it is alternating in nature.

Average electric current = 10 + 0 = 10 A.

The correct option is (c) 11.18 Ω

Explanation: The circuit is as shown in figure below.

Req = 5 + \(\frac{10(R_{eq}+5)}{10 + 5 + R_{eq}}\)

Or, \(R_{eq}^2 + 15R_{eq}\) = 5Req + 75 + 10Req + 50

Or, Req = \(\sqrt{125}\) = 11.18 Ω.

Right ANSWER is (a) 0.25 mA

For EXPLANATION: Since i = \(\frac{5}{10 × 10^{-3}}\) = 0.5 × 10^3 = 0.5 mA

As the switch is repeatedly close, then i(t) will be a square wave.

So average VALUE of electric current is (\(\frac{0.5}{2}\)) = 0.25 mA.

The correct choice is (a) 15.13 W

The explanation: In mesh aef, 8(I1 – I3) + 3(I1 – I2) = 15

Or, 11 I1 – 3 I2 – 8I3 = 15

In mesh EFD, 5(I2 – I3) + 2I2 + 3(I2 – I1) = 0

Or, -3 I1 + 10I2 – 5I3 = 0

In mesh abcde, 10I3 + 5(I3 – I2) + 8(I3 – I1) = 0

Or, -8I1 – 5I2 +23I3 = 0

Thus loop equations are,

11 I1 – 3 I2 – 8I3 = 15

-3 I1 + 10I2 – 5I3 = 0

-8I1 – 5I2 + 23I3 = 0

Solving by Cramer’s rule, I3 = current through the 10Ω resistor = 1.23 A

∴ Current through 10 Ω resistor = 1.23 A

Power loss (P) = \(I_3^2 r\) = (1.23)^2×10 = 15.13 W.

Correct choice is (d) L = 0.1 H, R = 2 Ω, C = 5 F

The explanation is: Dividing point impedance = R || sL || \(\frac{1}{sC} \)

= \(\BIG\{\frac{(R)(\frac{1}{sC})}{R + \frac{1}{sC}}\Big\}\) || sL

= \(\frac{\frac{R}{1+sRC}}{\frac{R}{1+sRC}+sL} \)

= \(\frac{sRL}{s^2 RLC+sL+R} \)

GIVEN that, Z(s) = \(\frac{0.2s}{s^2+0.1s+2} \)

∴ On comparing, we GET L = 0.1 H, R = 2 Ω, C = 5 F.

Correct choice is (a) 0.50 e^-25T mA

Easiest explanation: The CAPACITOR voltage VC (t) = VC (∞) – [VC (∞)-VC (0)]e^-t/RC

R = 20 || 20 = \(\FRAC{20×20}{20+20} = \frac{400}{40}\) = 10 kΩ

VC (∞) = 10 × \(\frac{20}{20+20}\) = 5 V

Given, VC (0) = 0

∴ VC (t) = 5 – (5-0)e^-t/10×4×10^(-6)×10^3

= 5(1 – e^-25t)

IC (t) = C\(\frac{dV_C (t)}{dt} = 4×10^{-6} \frac{d}{dt}5(1-e^{-25t})\)

= 4 × 10^-6 × 5 × 25e^-25t

∴ IL (t) = 0.50e^-2.5t mA.

The correct CHOICE is (d) 4 t^3

For explanation I would say: We KNOW that, I = \(\frac{1}{L} \int_0^t V \,dt\)

= 1\(\int_0^t 12 t^2 \,dt\)

= 4 t^3.

The CORRECT CHOICE is (c) Resistor

The best explanation: A linear circuit element does not change its value with voltage or CURRENT. The RESISTANCE is only ONE among the others does not change its value with voltage or current.

The correct answer is (d) Remains same

For explanation I would say: V / R RATIO is a CONSTANT R. If R is doubled then, ELECTRIC current will become HALF. So VOLTAGE across each resistor is same.

The correct option is (C) 2.8 Ω

Easiest explanation: Using Y-∆ TRANSFORMATION,

RAB = (9+9 || 6) || (9||6)

= (18 || 6) || (9 || 6)

=\(\LEFT(\FRAC{18×6}{18+6}\right) || \left(\frac{9×6}{9+6}\right)\)

= 4.5 || 3.6

= \(\frac{4.5×3.6}{4.5+3.6}\) = 2.8 Ω.

Right answer is (c) 0.25 W

The best I can EXPLAIN: For MAXIMUM POWER transfer to the load resistor RL, RL must be equal to 100Ω.

∴ Maximum power = \(\FRAC{V^2}{4R_L}\)

= \(\frac{10^2}{4×100} = \frac{100}{400}\) = 0.25 W.

Correct ANSWER is (d) 4.5 H

The best I can explain: We know that, M = K\(\sqrt{L_1 L_2}\)

Given that, L1 = 12 H, L2 = 5 H and k = 0.5

So, M = 0.5\(\sqrt{12 X 5}\)

= 0.5\(\sqrt{60}\)

= 0.5 X 7.75 = 3.875 H.

Right answer is (d) More than 1 Ω but LESS than 10 Ω 0 Ω

Best explanation: We know that impedance, Z= \( \sqrt{R^2 + (\frac{1}{ω^2+c^2})^2}\)

SINCE frequency is doubled, so \(\frac{1}{ω^2+c^2}\) BECOMES one-fourth but R2 remains the same. THUS the impedance cannot be EXACTLY measured but we can infer that the resistance is more than 1 Ω but less than 10 Ω.

Right option is (c) 625 mH

Explanation: We know that, M = k\(\SQRT{L_1 L_2}\)

GIVEN that, LEQ = 1500 mH and k = 0.2

Again, total inductance = L1 + L2 + 2M

Or, 2L + 2kL = 1500 mH

Or, L (2 + 2 X 0.2) = 1500

Or, L = 625 mH.

The correct answer is (C) V2 is leading V1 by 60°

For explanation I WOULD say: Given that, V1 = SIN (ωt + 30°) and V2 = cos (ωt)

Now, V2 can be WRITTEN as,

V2 = sin (ωt + 90°).

Hence, V2 is leading V1 by (90 – 30) = 60°.

The CORRECT choice is (c) 450 W, 2 W

The BEST I can explain: Let us suppose R = 1 Ω

Then, I1 = 32 A, I2 = 52 A and I3 = 72 A

Or, (I1 + I2 + I3)^2 R = (152)^2 R = 450 Ω.

Right choice is (d) 0.111 W

To explain: We KNOW that RMS current is,

I = 1 \( \int_0^1 (1T)^2\,dt\)

= \( \int_0^1 t^2\,dt\)

= \(\frac{1}{3}\) A

Now, power P = I^2R

= \(\frac{1}{9}\) X 1

= 0.111 W.

Right answer is (a) More than 50% of total energy

The explanation: The parallel combination of 30 Ω and 20 Ω is 12 Ω. Since 12 Ω and 10 Ω are in parallel, the 10 OHM plate draws more than 50% CURRENT and dissipates more than 50% energy.

Correct choice is (b) 2.5 J

Easiest EXPLANATION: We know that, Energy, E = 0.5 LI^2

Or, E = 0.5 X 5 X 1^2 = 2.5 J.

Correct answer is (b) Cannot be higher than source VOLTAGE

The explanation is: A practical voltage source has some resistance. Because of this resistance, some amount of voltage drop OCCURS across this resistance. HENCE, the terminal voltage cannot be higher than source voltage. However, if current is zero, then terminal voltage and source voltage are equal.

Correct ANSWER is (B) 1.67 Ω

To explain: We know that for STAR connection, REQ = \(\frac{R X R}{R+R+R}\)

GIVEN R = 5 Ω

So, REQ = \(\frac{5 X 5}{5+5+5}\)

= \(\frac{25}{15}\) = 1.67 Ω.

Correct CHOICE is (b) 60 C

To explain I would say: CHARGE Q= It

Given I = \(\frac{20}{200}\) = 0.1 A, t = 10 X 60 sec = 600 sec

So, Q = 0.1 X 600 = 60 C.

Right answer is (d) 0.0625 R

For explanation: Since diameter is double, area of cross-section is four TIMES and length is one-fourth.

It can be verified by the following EQUATION,

R2 = \(\FRAC{\frac{ρl}{4}}{4A}\)

= \(\frac{ρl}{16 A} = \frac{R}{16}\).

Right OPTION is (a) Decreases

To explain I would say: We KNOW that the temperature coefficient is,

Given by, α = \(\frac{α_0}{1 + α_0 t}\)

Since temperature is present at the denominator, so with INCREASE in temperature t, the denominator INCREASES and hence the fraction decreases.

So, temperature coefficient decreases.

The correct OPTION is (d) 4.2 W

To explain I would say: Equivalent resistance of the circuit is = [{(3 + 2) || 5} + 10]

= (2.5 + 10) = 12.5 Ω

Total current drawn by the circuit is IT = \(\frac{50}{12.5}\) = 4 A

Current in 3 Ω resistor is I3 = IT ×\(\frac{5}{5+5} = \frac{4 × 5}{10}\) = 2 A

VTH = V3 = 3 × 2 = 6V

RTH = RAB = [(2 + 5) || 3] = 2.1 Ω

For maximum power transfer RL = RTH = 2.1 Ω

∴ Current drawn by RL is IL = \(\frac{6}{2.1+2.1} = \frac{6}{4.2}\) = 1.42 A

∴ Power DELIVERED to the LOAD = \(I_L^2 R_L\)

= (1.42)^2(2.1) = 4.2 W.

The correct CHOICE is (d) 621 W

To EXPLAIN: RL = \(\sqrt{R_{TH}^2+X_{TH}^2}\)

= \(\sqrt{3^2+4^2}\) = 5

Now, 110 = (6 + j8 + 5) I1 + 5I2

And 90 = (6 + j8 + 5) I2 = 5I1

∴ I1 = 5.5 – 2.75j and I2 = 4.5 – 2.2j

Total current in RL = I1 + I2 = (10 – 4.95j) A = 11.15 A

∴ Power absorbed by RL = I^2R

= 11.15^2 × 5 = 621 W.

Correct CHOICE is (a) The circuit follows Reciprocity Theorem

Easiest explanation: Let us consider this circuit,

Equivalent Resistance, 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, I1 = \(\frac{3 × 60}{30+60}\) = 2 A.

Again, let us consider this circuit,

Equivalent resistance, REQ = [[20 || 60] + 30]

= \(\Big[\frac{20 × 60}{20+60} + 30\Big]\)

= [15 + 30] = 45 Ω

Total current = \(\frac{120}{45}\) = 2.67 A

Current through the 20Ω RESISTOR is, I2 = \(\frac{2.67 × 60}{60+20}\) = 2 A

Since I1 = I2, the circuit follows Reciprocity Theorem.

The correct CHOICE is (b) 3 Ω

The explanation: I + 0.9 = 10 I

Or, I = 0.1 A

VOC = 3 × 10 I = 30 I

Or, VOC = 3 V

Now, ISC = 10 I = 1 A

Rth = 3/1 = 3 Ω.

The correct choice is (d) 2 A

To explain I would SAY: Equivalent RESISTANCE, REQ = [[20 || 60] + 30]

= \(\Big[\frac{20 × 60}{20+60} + 30\Big]\)

= [15 + 30] = 45 Ω

Total current = \(\frac{120}{45}\) = 2.67 A

Current through the 20Ω RESISTOR is, I20Ω = \(\frac{2.67 × 60}{60+20}\) = 2 A.