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The correct option is (d) 0.041 Hz

Best explanation: IEQ = L1 + L2 + 2M

LEQ = 10 + 20 + 2 × \(\frac{1}{20}\) = 30.1 H

∴ FO = \(\frac{1}{2π\sqrt{LC}}\)

= \(\frac{1}{2π\sqrt{30.1 × 0.5}}\)

= 0.041 Hz.

Right answer is (c) 3

Best EXPLANATION: The total number of independent loop EQUATIONS are given by L = B – N + 1 where,

L = number of loop equations

B = number of BRANCHES = 10

N = number of nodes = 8

∴ L = 10 – 8 + 1 = 3.

Right ANSWER is (c) 325 V

The best I can EXPLAIN: Loop CURRENT I1 = \(\frac{50}{20}\) = 2.5 A

I2 = \(\frac{100}{20}\) = 5 A

VAB = (50) (2.5) + 100 + (5) (20)

 = 125 + 100 + 100

 = 325 V.

The correct CHOICE is (d) 45°

The best explanation: VI cos θ = 0.25 or I cos θ = 0.25

Or, Z cosθ = 2

Or, \(\frac{V}{I}\) cos⁡θ = 2

Or, cos θ = \(\frac{1}{\sqrt{2}}\)

So, from the above EQUATIONS, cos θ = 0.707 and θ = 45°.

The CORRECT answer is (b) 3.1 F

To elaborate: CCB = \(\left(\frac{C_2 C_3}{C_2+ C_3}\right)\) + C5 = 7.5 F

Now, CAB = \(\left(\frac{C_1 C_{CB}}{C_1 + C_{CB}}\right)\) + C6 = 8 F

CXY = \(\frac{C_{AB}× C_4}{C_{AB} + C_4}\) = 3.1 F.

Correct choice is (c) 3 V

The explanation: When R = ∞, V = 5v,

Then, Voc = 5V and the CIRCUIT is open

When R = 0, I = 2.5A

Then, ISC = 2.5 and the circuit is short circuited.

So, Req = \(\frac{V_{OC}}{I_{SC}}\)

= \(\frac{5}{2.5}\) = 2 Ω

Hence the voltage across 3 Ω is 3 volt.

Correct option is (a) 10 mH

For EXPLANATION: We know that if an inductor L is connected in delta, then the equivalent star of each ARM = \(\FRAC{L X L}{L+L+L}\)

Given that, L = 30 mH

= \(\frac{30 X 30}{30+30+30}\)

= \(\frac{900}{90}\) = 10 mH.

Right CHOICE is (b) 6 Ω

For EXPLANATION I WOULD say: The RESISTANCE of parallel combination is given by,

Req = \(\frac{40}{3} \) – 10 = 3.33 Ω

Or, \(\frac{1}{3.33} = \frac{1}{12} + \frac{1}{15} + \frac{1}{15}\)

Or, R = 6 Ω.

Correct ANSWER is (b) VTH = 0.5 V, RTH = 0.75 Ω

To explain: VTH = \(\frac{1 X 20}{40}\)X 1 = 0.5 V

Also, RTH = \(\frac{1 X 30}{40}\) = 0.75 Ω.

The CORRECT CHOICE is (a) 1 – e^-t/RC

Best EXPLANATION: At t = 0, VC = 0 and at t = ∞, Vc = 1.

This condition can be satisfied only by (1 – e^-t/RC).

The CORRECT OPTION is (d) 3 V

To explain: The average value of the waveform = \(\frac{2 X AREA \,of \,1st \,triangle\, + \,Area \,of\, 2nd \,triangle}{π}\)

= \(\frac{2 X \frac{π}{3} X \frac{1}{2} X 6 + 6(\frac{π}{2} – \frac{π}{3})}{π}\)

= \(\frac{2π + π}{π}\) = 3 VOLT.

The correct choice is (B) 6.7 V

To explain I would say: We know that, during DISCHARGE of capacitor,

VC = VR

Now, VR = 0.67 X 10 = 6.7 V

So, VC= 6.7 V.

The CORRECT ANSWER is (d) 4 A

Best explanation: At steady state, the circuit BECOMES,

∴ The current through the voltage SOURCE V= 5 – 1 = 4 A.

The correct CHOICE is (c) 20 A

The EXPLANATION is: CURRENTS in resistance and inductance are out of phase by 90°.

Hence, I = \(I_1^2 + I_2^2\)

Or, I = [12^2 + 16^2]^0.5

Or, I = \(\sqrt{144+256} = \sqrt{400}\)

= 20 A.

Right answer is (C) V (t) = R1i1 + L1 \(\frac{di_1}{dt}\) – M \(\frac{di_2}{dt}\)

To explain: We know that, in general, the KVL is of the form V (t) = R1i1 + L1 \(\frac{di_1}{dt}\) + M \(\frac{di_2}{dt}\)

But here, M term is NEGATIVE because i1, is entering the dotted terminal and i2, is leaving the dotted terminal.

So, V (t) = R1i1 + L1 \(\frac{di_1}{dt}\) – M \(\frac{di_2}{dt}\).

Right choice is (a) 14.7 A

Explanation: Using KVL, 100 = \(R\frac{dq}{DT} + \frac{Q}{C}\)

100 C = \(RC\frac{dq}{dt}\) + q

Or, \(\int_{q_o}^q \frac{dq}{100C-q} = \frac{1}{RC} ∫_0^t \,dt\)

100C – q = (100C – qo)e^-t/RC

I = \(\frac{dq}{dt} = \frac{(100C – q_o)}{RC} e^{-1/1}\)

∴ e^-t/RC = 40e^-1 = 14.7 A.

The correct choice is (b) 6 Ω and 0.833 A

The explanation is: We, draw the Norton equivalent of the left SIDE of xx’ and SOURCE transformed right side of yy’.

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

∴ RN = 8 || (16 + 8)

= \(\frac{8×24}{8+24}\) = 6 Ω

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

The correct answer is (b) 5 A

For explanation: At steady state, the CIRCUIT becomes,

∴ The CURRENT through R3 = \(\frac{5}{1}\) = 5 A.

Correct CHOICE is (c) 3.18 MH

Explanation: VL = VC = 2 VR

∴ Q = \(\frac{V_L}{V_R}\)= 2

But we know, Q = \(\frac{ωL}{R} = \frac{1}{ωCR}\)

∴ 2 = \(\frac{2πf × L}{5}\)

Or, L = 3.18 mH.

The correct option is (a) 6 V

The BEST I can explain: For finding the Thevenin Equivalent CIRCUIT across A-B we remove the 5 Ω resistor.

Then, I = \(\frac{10+50}{15}\) = 4 A

VOC = 50 – (10×4) =10 V

And REQ = \(\frac{10×5}{10+5} = \frac{10}{3}\) Ω

Current I1 = \(\frac{10}{10/3+5} = \frac{6}{5}\)

Hence, VAB = \(\frac{6}{5 × 5}\) = 6 V.

Right choice is (a) 0

For EXPLANATION: DYNAMIC resistance of the TANK circuit, ZDY = \(\frac{L}{RLC}\)

But given that RL = 0

So, ZDY = \(\frac{L}{0XC}\) = ∞

Therefore current through circuit, I = \(\frac{V}{∞}\) = 0

∴ VD = 0.

The correct choice is (d) 12 Ω

Explanation: Using KVL in loop 1, we get, 100 – 14I – 30 = 0

Or, I = \(\frac{70}{14}\) = 5 A

Then, VP – VQ = 14I – (15-I).1

= 70 – 10 = 60 V

∴ R = \(\frac{60}{10-I} = \frac{60}{5}\) = 12 Ω.

The correct option is (c) V = (L1 + L2 + L3 + 2M12 – 2M23 – 2M13) \(\FRAC{di}{dt}\)

To explain: M12 is positive while M23 and M13 are negative because of dots SHOWN in figure.

So, the KVL equation is given by,

V = (L1 + L2 + L3 + 2M12 – 2M23 – 2M13) \(\frac{di}{dt}\).

The CORRECT option is (d) 12 A

The best I can explain: We can infer from the CIRCUIT,

A2 = \(\sqrt{13^2 – 5^2}\)

Or, A2 = \(\sqrt{169 – 25}\)

Or, A2 = \(\sqrt{144}\)

Or, A2 = 12 A.

The correct choice is (b) -25.23 V

The explanation is: \(\frac{V_A}{30} + \frac{V_A-40}{12} + \frac{V_A-V_B}{8}\) = 0

Or, 29 VA – 15 VB = 400

Also, \(\frac{V_B-V_A}{8} + \frac{V_B-120}{8}\) + 6 = 0

Or, VA = 65.23 V, VB = 99.44 V

V1 = 40-65.23 = -25.23 V.

Right answer is (d) -90°

For explanation: NET VOLTAGE applied to the circuit is 200∠0° V

I1 = \(\frac{200∠0°}{10.0}\)

= 20∠0° = 20

I2 = \(\frac{200∠0°}{10∠90°}\)

= 20∠-90° = -j20

I = I1 + I2 = 20(1-j) = 20\(\SQRT{2}\)∠45°

Voltage V1 = 100(1+j)

= 100\(\sqrt{2}\)∠45°

∴ Required phase angle = -45° – 45° = -90°.

The correct CHOICE is (c) Capacitance

Explanation: We know that, when a current pulse is applied to a capacitor, the VOLTAGE will have a WAVEFORM which rises linearly and then becomes constant TOWARDS the end of pulse. Hence, the ELEMENT is a capacitor.

Correct option is (d) 500 Hz

For explanation I would say: We know that, maximum voltage across CAPACITANCE occurs at a frequency slightly less than RESONANT frequency.

Here, given that resonant frequency = 550 Hz.

So, out of the given options 500 is the LOWEST nearest integer to 550.

Hence, 975 Hz.

Correct answer is (B) INCREASE

The EXPLANATION: We know that the specific gravity of ELECTROLYTE is HIGHEST when battery is fully charged and is lowest when discharged. So, the specific gravity of the electrolyte will increase when a lead acid battery is being charged.

The correct CHOICE is (B) 6 – 3e^-t

The EXPLANATION is: From the figure, we can infer that,

I(t) = \(\frac{12}{12} \left(1 – \frac{2}{4}e^{-t}\right)\)

= 6 – 3e^-t.

Right CHOICE is (d) \(\frac{V_S RR_2}{R_1 R_2 + R_1 R + RR_2}\)

Explanation: EFFECTIVE resistance of R and Rm is

Req = \(\frac{RR_m}{R + R_m}\)

Therefore the reading is \(\frac{V}{R_1 + \frac{R}{R + R_2}}\BIG[\frac{RR_2}{R+R_2}\Big]\)

= \(\frac{V_S RR_2}{R_1 R_2 + R_1 R + RR_2}\).

The CORRECT OPTION is (C) 55 μC and 275 μC respectively

For explanation I WOULD say: Q1 = 0.5 x 10^-6 x 110

= 55 μC.

Also, Q2 = 2.5 x 10^-6 x 110

= 275 μC.

Right option is (d) 40 Ω

Explanation: The current in the branch CD is ZERO if the POTENTIAL difference in the branch CD is zero.

That is, VC = VD

Or, V10 = VC = VD = VA × \(\frac{10}{15}\)

VR = VA × \(\frac{R}{20+R} \)

And V10 = VR

∴ VA × \(\frac{10}{15}\) = VA × \(\frac{R}{R+20} \)

∴ R = 40 Ω.

The correct answer is (b) -3.41 V

The best explanation: APPLYING KCL to node A, \(\FRAC{V_A-10}{10} + \frac{V_A}{20} + \frac{V_A}{7}\) = 0

Or, VA (0.1 + 0.05 + 0.143) = 1

Or, VA = 3.41 V

The voltage across the 2 Ω resistor due to 10 V source is V2 = \(\frac{V_A}{7} × 2\) = 0.97 V

V2Ω due to 20 V source, \(\frac{V_A}{10} + \frac{V_A}{20} + \frac{V_A-20}{7}\) = 0

Or, 0.1 VA + 0.05VA + 0.143VA = 2.86

∴ VA = \(\frac{2.86}{0.293}\) = 9.76 V

V2Ω = \(\frac{V_A-20}{7}\) × 2 = -2.92 V

The CURRENT in 2 Ω resistor = 2 × \(\frac{5}{5+8.67}\)

= \(\frac{10}{13.67}\) = 0.73 A

The voltage across the 2 Ω resistor = 0.73 × 2 = 1.46 V

V2Ω = 0.97 – 2.92 -1.46 = -3.41 V.

The correct choice is (a) 125 V

For explanation I would say: Current through the 18 Ω resistance = \(\frac{90}{18}\) = 5 A

Equivalent resistance of 3 Ω and 7 Ω banks = \(\frac{3 × 6}{3 + 6}\) = 2 Ω

Since, this 2 Ω resistance is in series with 18 Ω resistance, THEREFORE total resistance = 18 + 2 = 20 Ω

This 20 Ω resistance is in parallel with 5 Ω resistance = \(\frac{5 × 20}{5 + 20}\) = 4 Ω

Hence, total resistance of the circuit = 1 + 4 = 5 Ω

Current through this BRANCH = 5 A

∴ Voltage across dc= 5 × 20 = 100 V

Hence current through 5 Ω resistance = \(\frac{100}{5}\) = 20 A

∴ Total current = 20 + 5 = 25 A

Since, total resistance of the circuit is 5 Ω therefore, voltage E = 25 × 5 = 125 V.

The correct answer is (d) 2.5 and 2.5 V

To explain: I1 = \(\FRAC{V_1-V_2}{2}\)

APPLYING KCL at node 1, 5 = \(\frac{V_1}{1} + \frac{V_1-V_2}{2} + V_1 \)

10 = 2V1 + V1 – V2 + 2V1

Or, 10 = 5V1 – V2

KCL at node 2, \(\frac{V_1-V_2}{2}\) + V1+ 2I1= V2

∴ 1.5 V1 – V2= 0

∴ V1 = V2= 2.5 V.

Right option is (B) 2.5 A

Easiest explanation: Applying KCL, we get, I1 + 5 = I2 + I3

∴ \(\FRAC{10 – V_1}{1} + 5 = \frac{V_1}{2} + \frac{V_2}{6}\)

∴ 30 – 3V1 + 15 = 3V1 + V2

∴ 6 V1 + V2 = 45

From voltage source, V2 – V1 = 10

Now, 7 V1 = 35, V1 = 5 V

And V2 = 15 V

∴ IX = \(\frac{V_2}{6} = \frac{15}{6}\) = 2.5 A.

Right choice is (b) 1 V SOURCE in series with 2.4 Ω resistor

The best I can explain: APPLYING Thevenin’s Theorem REQ = 6 || 4

= \(\frac{6 × 4}{6 + 4}\)

= \(\frac{24}{10}\) = 2.4 Ω

VAB = 10 – 6 × (\(\frac{15}{10}\)) = 1 V.

The CORRECT option is (b) RA = 2.5 Ω and RB = 4.5 Ω

To explain I WOULD SAY: GIVEN RA + RB = 7 with C open

RB + RC = 12 with A as open

RA + RC = 10 with B open

Then, RA + RB + RC = \(\frac{29}{2}\) = 14.5

Hence, RA = 2.5 Ω, RB = 4.5 Ω and RC = 7.5 Ω.

Correct choice is (d) -10 cos(ωt + 36.87) mA

To explain I would SAY: Applying KCL, we get, I1 + I2 + I3 = 0

∴ -6 sin(ωt) + 8 cos(ωt) + I3 = 0

∴ I3 = 6 sin (ωt) – 8 cos (ωt)

= 10[sin (ωt).sin (36.86) – cos (ωt) cos (36.86)]

=-10[cos (ωt) cos (36.86) – sin (ωt) sin (36.86)]

= -10 cos (ωt + 36.86)

[As, cos (A+B) = cosA.cosB – sinA.sinB].

Right answer is (d) 12 Ω

Best explanation: By KCL,

∴ \(\frac{V_P-40}{1} + \frac{V_P-100}{14} + \frac{V_P}{2}\) = 0

Or, 22 VP = 660

∴ VP = 30 V

Potential difference between NODE x and y = 60 V

∴ -I – 5 + \(\frac{40-30}{I}\) = 0

Or, I = 5 A

∴ R = \(\frac{60}{5}\)= 12 Ω.

Correct ANSWER is (B) 100 ∠73.72°

Easy explanation: GIVEN A = 3 + J1

So, 3 + j1 = 10∠18.43°

Or, 3 + j1 = (10)^4 ∠4 X 18.43°

= 100∠73.72°.

The correct ANSWER is (a) 2.5, 5 and 5

Easiest explanation: R1 = \(\frac{R_B R_C}{R_A+R_B+R_C} = \frac{100}{40}\) = 2.5 Ω

R2 = \(\frac{R_A R_C}{R_A+R_B+R_C} = \frac{200}{40}\) = 5 Ω

R3 = \(\frac{R_B R_A}{R_A+R_B+R_C} = \frac{200}{40}\) = 5 Ω.

Correct option is (a) – 3V

The explanation: By APPLYING KVL, I = 1 A

VAB – 2 × 1 + 5 = 0

Or, VAB = -3 V.

The correct answer is (a) High RELATIVE permittivity

The explanation: Relative permittivity is for ideal dielectric which is AIR. ACHIEVING such a precise dielectric is very difficult.

Low relative permittivity will lead to low VALUE of capacitance.

High relative permittivity will lead to a HIGHER value of capacitance.

The correct option is (b) \(\SQRT{3}\)∠-30°

Easy explanation: Given that, ‘a’ = 1 ∠120°

So, 1 – a = 1 – 1∠120°

= 1 + 0.5 – J 0.866

= 1.5 – j 0.866

= 3∠-30°.