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The correct option is (b) 6.67%

The best explanation: RT = \(\frac{V_T}{I_T}\)

VT = 200 V,IT = 10 A

So, RT = 20 kΩ

Resistance of VOLTMETER,

RV = 1000 × 300 = 300 kΩ

Voltmeter is in parallel with unknown RESISTOR,

RX = \(\frac{R_T R_V}{R_T – R_V} = \frac{20 ×300}{280}\) = 21.43 kΩ

Percentage error = \(\frac{Actual-Apparent}{Actual}\) × 100

= \(\frac{21.43-20}{21.43}\) × 100 = 6.67%.

Right CHOICE is (c) 25.0 Ω

The best explanation: Ish Rsh = IM Rm

Ish = I – Im or, \(\frac{I}{I_m} – 1 = \frac{R_m}{R_{SH}}\)

Now, m = \(\frac{I}{I_m}\)

Or, m – 1 = \(\frac{R_m}{R_{sh}}\)

∴Rsh = 25 Ω.

The CORRECT OPTION is (d) 6.67%

Easiest explanation: ERROR in 10 kΩ resistance = 10 × \(\FRAC{5}{100}\) = 0.5 kΩ

Error in 5 kΩ resistance = 5 × \(\frac{10}{100}\) = 5 kΩ

Total MEASUREMENT resistance = 10 + 0.5 + 5 + 0.5 = 16 kΩ

Original resistance = 10 + 5 = 15 kΩ

Error = \(\frac{16-15}{15}\) × 100 = \(\frac{1}{15}\) × 100 = 6.67%.

Right CHOICE is (a) 0.133 Ω

To elaborate: Inverse Hybrid parameter g11 is GIVEN by, g11 = \(\FRAC{I_1}{V_1}\), when I2 = 0.

Therefore short circuiting the TERMINAL Y-Y’, we get,

 V1 = I1 ((5||5) + 5)

= I1 \(\left(\left(\frac{5×5}{5+5}\right) + 5\right)\)

= 7.5I1

∴ \(\frac{I_1}{V_1}= \frac{1}{7.5}\) = 0.133 Ω

Hence g11 = 0.133 Ω.

Right answer is (b) 0.5 Ω

Explanation: Hybrid PARAMETER h21 is given by, h21 = \(\FRAC{I_2}{I_1}\), when V2 = 0.

Therefore short circuiting the terminal Y-Y’, and applying Kirchhoff’s LAW, we get,

-50 I2 – (I2 – I1)50 = 0

Or, -I2 = I2 – I1

Or, -2I2 = -I1

∴\(\frac{I_2}{I_1} = \frac{1}{2}\)

HENCE h21 = 0.5 Ω.

Right CHOICE is (a) 75 Ω

The best I can EXPLAIN: Hybrid parameter h11 is given by, h11 = \(\frac{V_1}{I_1}\), when V2=0.

Therefore short circuiting the TERMINAL Y-Y’, we get,

 V1 = I1 ((50||50) + 50)

= I1 \(\left(\left(\frac{50×50}{50+50}\right) + 50\right)\)

= 75I1

∴ \(\frac{V_1}{I_1}\) = 75.

Hence h11 = 75 Ω.

Right answer is (d) 240 V

Easiest explanation: In order for 600 C charges to be delivered to 100 V source, the electric current must be in REVERSE clockwise direction.

Now, I = \(\frac{dQ}{DT}\)

= \(\frac{600}{60}\) = 10 A

Applying KVL we get

V1 + 60 – 100 = 10 × 20 ⇒ V1 = 240 V.

The correct option is (b) 0.05 J

Explanation: E = \(\frac{1}{2}\)CV^2

= 5 X 10^-6 X 100^2

= 0.05 J.

Correct CHOICE is (c) 111 ± 5.55 Ω

For explanation I would say: R1 = 36 ± 5% = 36 ± 1.8 Ω

R2 = 75 ± 5% = 75 ± 3.75 Ω

∴ R1 + R2 = 111 ± 5.55 Ω.

Right choice is (c) 10 A

The best explanation: Average dc electric CURRENT = 10 A.

Average ac electric current = 0 A SINCE it is ALTERNATING in nature.

Average electric current = 10 + 0 = 10 A.

Correct OPTION is (c) 0.125 mA

The explanation: Since i = \(\frac{5}{10 × 2 X 10^{-3}}\) = 0.25 × 10^-3 = 0.25 mA.

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

So average value of electric current is (\(\frac{0.25}{2}\)) = 0.125 mA.

The correct OPTION is (c) 11.18 Ω

To EXPLAIN I WOULD say: The circuit is as shown in figure below.

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

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

Or, \(R_{eq} = \sqrt{125}\) = 11.18 Ω.

The correct option is (d) 4 t^3

Best explanation: We know that, I = \( \frac{1}{L} \int_0^t V \,DT\)

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

= 4 t^3.

Correct answer is (C) Resistor

To explain I WOULD SAY: A LINEAR circuit element does not change their 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

To explain: The voltage/resistance RATIO is a constant (say K). If K is DOUBLED then, electric current will become half. So voltage across each resistor remains same as was initially.

Correct option is (d) Open circuit OUTPUT admittance

Best explanation: We know that, h22 = \(\frac{I_2}{V_2}\), when I1 = 0.

Since the current in the first loop is 0 when the RATIO of the current and voltage in SECOND loop is MEASURED, therefore the parameter h12 is called as Open circuit output admittance.

Correct OPTION is (d) h11 h22 – h12 h22 = 1

The EXPLANATION: We know that, I1 = y11 V1 + y12 V2 ……… (1)

I2 = y21 V1 + y22 V2 ………. (2)

And, V1 = h11 I1 + h12 V2 ………. (3)

I2 = h21 I1 + h22 V2 ……….. (4)

Now, (3) and (4) can be rewritten as,

I1 = \(\frac{V_1}{h_{11}}– \frac{h_{12} V_2}{h_{11}}\)………. (5)

And I2 = \(\frac{h_21 V_1}{h_11}+ \left(- \frac{h_{21} h_{12}}{h_{11}} + h_{22}\right) V_2\) ………. (6)

Therefore using the above 6 equations in representing the hybrid parameters in terms of the Y parameters and applying ∆Y = 0, we get,

h11 h22 – h12 h22 = 1 [hence proved].

The CORRECT choice is (B) 0.5 Ω

Easiest EXPLANATION: Hybrid PARAMETER h21 is given by, h21 = \(\frac{I_2}{I_1}\), when V2 = 0.

Therefore short circuiting the terminal Y-Y’, and applying Kirchhoff’s law, we GET,

-5 I2 – (I2 – I1)5 = 0

Or, -I2 = I2 – I1

Or, -2I2 = -I1

∴\(\frac{I_2}{I_1} = \frac{1}{2}\)

 Hence h21 = 0.5 Ω.

The CORRECT option is (c) Z22 = 30 Ω

The explanation: z11 = \(\frac{V_1}{I_1}= \frac{(20+5)I_1}{I_1}\) = 25Ω

V0 = \(\frac{20}{25}\)V1 = 20 I1

-V0 – 4I2 + V2 = 0

Or, V2 = V0 + 4I1 = 20I1 + 4I1 = 24 I1

Or, z21 = \(\frac{V_2}{I_1}\) = 24 Ω

V2 = (10+20) I2 = 30 I2

Or, z22 = \(\frac{V_2}{I_1}\) = 30 Ω

V1 = 20I2

Or, Z12 = \(\frac{V_1}{I_2}\) = 20 Ω

∴ [z] = [25:20; 24:30] Ω.

The correct OPTION is (c) 440 V

Easy explanation: We know that,

PEAK voltage rating = 2 (rms voltage rating)

Given that the RMS voltage rating = 220 V

So, the Peak Voltage Rating = 2 X 220 V

= 440 V.

The correct choice is (c) z22 = -1.775 – j5.739 Ω

Easiest EXPLANATION: z1 = \(\frac{12(J10)}{12+j10-j5} = \frac{j120}{12+j5}\)

z2 = \(\frac{j60}{12+j5}\)

Z3 = \(\frac{50}{12+j5}\)

z12 = Z21 = z2 = \(\frac{(-j60)(12-j5)}{144+25}\) = -1.775 – j4.26

z11 = z1 + z12 = \(\frac{(j120)(12-j5)}{144+25}\) + z12 = 1.775 + j4.26

z22 = z3 + z21 = \(\frac{(50)(12-j5)}{144+25}\) + z21 = 1.7758 – j5.739

∴ [z] = [1.775 + j4.26; -1.775 – j4.26; -1.775 – j4.26; 1.775 – j5.739] Ω.

The correct answer is (c) z22 = -J4 Ω

Explanation: z12 = j6 = z21

Z11 – z12 = 4

Or, z11 = z12 + 4 = 4 + j6 Ω

And z22 – z12 = -j10

Or, z22 = z12 + -j10 = -j4 Ω

∴ [Z] = [4+j6:j6; j6:-j4] Ω.

The correct option is (b) z22 = 2.773 Ω

To explain I WOULD SAY: z11 = \(\frac{V_1}{I_1}\)= 2 + 1 || [2+1 || (2+1)]

z11 = 2 + 1 || (2 + \(\frac{3}{4}\)) = 2 + \(\frac{1×\frac{11}{4}}{1+\frac{11}{4}} = 2 + \frac{11}{15}\) = 2.733

I0 = \(\frac{1}{1+3}\) I’0 = \(\frac{1}{4}\) I’0

And I’0 = 1 + \(\frac{11}{4}\)I1 = \(\frac{4}{15}\) I1

Or, I0 = \(\frac{1}{4} × \frac{4}{5} I_1 = \frac{1}{15} I_1\)

Or, V2 = I0 = \(\frac{1}{15} I_1\)

z21 = \(\frac{V_2}{I_1} = \frac{1}{15}\) = z12 = 0.0667

z22 = \(\frac{V_2}{I_2}\)= 2+1 || (2+1||3) = z11 = 2.733

∴ [z] = [2.733:0.0667; 0.0667:2.733] Ω.

Right answer is (c) z22 = 1.667 Ω

For explanation: Z11 = \(\frac{V_1}{I_1}\)= 1 + 6 || (4+2) = 4Ω

I0 =\(\frac{1}{2} I_1 \)

V2 = 2I0 = I1

z21 = \(\frac{V_2}{I_1}\) = 1Ω

z22 = \(\frac{V_2}{I_2}\)= 2 || (4+6) = 1.667Ω

So, I’0 = \(\frac{2}{2+10}I_2 = \frac{1}{6}I_2 \)

V1 = 6I’0 = I2

z12 = \(\frac{V_1}{I_2}\)= 1Ω

Hence, [Z] = [4:1; 1:1.667] Ω.

Correct ANSWER is (c) 3 A and 0

Easiest explanation: I(s) = \(\frac{6}{s+1} – \frac{3}{s+0.5}\)

Or, I(t) = 6 e^-t – 3 e^-0.5t

Putting, t = 0, we get, I(0) = 3A

Putting t = ∞, we get, I (∞) = 0.

Correct answer is (C) 9 Ω

The EXPLANATION is: We KNOW that,

When R = 0, circuit current = \(\frac{V}{30}\) A

And Power dissipated = \(\frac{V^2}{30}\) Watts.

This is the maximum possible value which occurs for R = 0 Ω.

The correct choice is (d) Z11 = 1.775 + j4.26 Ω

To ELABORATE: z1 = \(\frac{12(J10)}{12+j10-j5} = \frac{j120}{12+j5}\)

z2 = \(\frac{j60}{12+j5}\)

Z3 = \(\frac{50}{12+j5}\)

z12 = z21 = z2 = \(\frac{(-j60)(12-j5)}{144+25}\) = -1.775 – j4.26

z11 = z1 + z12 = \(\frac{(j120)(12-j5)}{144+25}\) + z12 = 1.775 + j4.26

z22 = z3 + z21 = \(\frac{(50)(12-j5)}{144+25}\) + z21 = 1.7758 – j5.739

∴ [z] = [1.775 + j4.26; -1.775 – j4.26; -1.775 – j4.26; 1.775 – j5.739] Ω.

Correct CHOICE is (a) z21 = 0.0667 Ω

For EXPLANATION: z11 = \(\frac{V_1}{I_1}\)= 2 + 1 || [2+1 || (2+1)]

z11 = 2 + 1 || (2 + \(\frac{3}{4}\)) = 2 + \(\frac{1×\frac{11}{4}}{1+\frac{11}{4}} = 2 + \frac{11}{15}\) = 2.733

I0 = \(\frac{1}{1+3}\) I’0 = \(\frac{1}{4}\) I’0

And I’0 = 1 + \(\frac{11}{4}\)I1 = \(\frac{4}{15}\) I1

Or, I0 = \(\frac{1}{4} × \frac{4}{5} I_1 = \frac{1}{15} I_1\)

Or, V2 = I0 = \(\frac{1}{15} I_1\)

z21 = \(\frac{V_2}{I_1} = \frac{1}{15}\) = z12 = 0.0667

z22 = \(\frac{V_2}{I_2}\)= 2+1 || (2+1||3) = z11 = 2.733

∴ [z] = [2.733:0.0667; 0.0667:2.733] Ω.

Correct ANSWER is (b) 0

The EXPLANATION: The CURRENT i (t) is given by,

i = 25 sin 314.16 t and \(\FRAC{di}{dt}\) = 250 X 314.16 cos⁡ωt

Now, at t = 0.005, I = 25 X 314.16 cos (314.16 X 0.005)

= 0.

The CORRECT option is (a) 0.3 V

Explanation: The emf is given by,

V = M\(\frac{DI}{dt}\)

= \(\frac{3}{1000}\) X 100 = 0.3 V

Hence, the emf induced in coil Y is given by 0.3 V.

The correct choice is (a) Mmf for IRON and air gap are equal

For EXPLANATION: We know that, MMF for air = \(\FRAC{B}{4π X 10^{-7}}\) X 10

Where B is the magnetic field intensity.

Also, MMF for iron = \(\frac{B X 100}{200(4π X 10^{-7})}\)

= \(\frac{B X 0.5}{4π X 10^{-7}}\).

Correct option is (B) 6 V in series with 1.2 Ω RESISTANCE

Easiest explanation: The Thevenin equivalent voltage is given by,

VTH = \(\frac{10 X 3}{5}\)

= 6 V

And the Thevenin equivalent Resistance is given by,

RTH = \(\frac{3 X 2}{5}\) = 1.2 Ω.

The correct choice is (b) SYMMETRICAL networks

Best explanation: A symmetrical NETWORK can be split into TWO HALVES. So the z parameters of the network are symmetrical as well as reciprocal of each other. Hence Barletts Bisection THEOREM is applicable to Symmetrical networks.

The CORRECT answer is (c) 10800 W

To explain: Power dissipated instantaneously across the blades of the switch is GIVEN by,

Power, P = \(\FRAC{Energy}{Time}\)

Given that, Energy = 108 J and time = 10 X 10^-3

So, P = \(\frac{108}{10 X 10^{-3}}\) = 10800 W.

The correct choice is (b) 2 A

To elaborate: At t = ∞ the circuit has effectively two 6Ω RESISTANCES in parallel.

So, REQ = \(\frac{6 X 6}{6 + 6}\)

= \(\frac{36}{12}\) = 3 Ω

Given voltage = 6 V

So, CURRENT = 2 A.

The correct answer is (c) 0.2 nF

For explanation I WOULD SAY: K1 = \(\frac{R_2}{R_1}\)

= \(\frac{10^4}{20}\) = 5 X 10^2 = 500 Ω

And \(\frac{L_1}{L_2}= \frac{kω}{k_1}\)

Or, \(\frac{kω}{k_1} =\frac{L_1}{L_2}X k_1\)

Or, \(\frac{10^{-2}}{10^{-4} X 5 X 10^2}\) = 5 X 10^4

Or, C2 = \(\frac{C_1}{kω.k_1} = \frac{0.5 X 10^{-2}}{5 X 10^4 X 500}\)

= 0.02 X 10^-8 = 0.2 nF.

The correct answer is (d) z22 = \(\left(g_{22} – \FRAC{g_{21} g_{12}}{g_{11}}\right)\)

EASY explanation: We know that, V1 = z11 I1 + z12 I2 ……… (1)

V2 = Z21 I1 + z22 I2 ………. (2)

And, I1 = G11 V1 + g12 I2 ………. (3)

V2 = g21 V1 + g22 I2 ……….. (4)

Now, (3) and (4) can be rewritten as,

V1 = \(\frac{I_1}{g_{11}}– \frac{g_{12}}{g_{11}} I_2\) ………… (5)

And V2 = \(\left(g_{22} – \frac{g_{21} g_{12}}{g_{11}}\right) I_2 – \frac{g_{21} I_1}{g_{11}}\)……….. (6)

∴ Comparing (1), (2) and (5), (6), we GET,

z11 = \(\frac{1}{g_{11}} \)

z12 = – \(\frac{g_{12}}{g_{11}} \)

z21 = – \(\frac{g_{21}}{g_{11}} \)

z22 = \(\left(g_{22} – \frac{g_{21} g_{12}}{g_{11}}\right)\).

The correct CHOICE is (c) Four times

To elaborate: Since diameter is DOUBLED, area of cross-section becomes four times. Current carrying CAPACITY is proportional to area of cross-section.

Correct answer is (b) z12 = – \(\frac{g_{12}}{g_{11}} \)

To explain: We know that, V1 = z11 I1 + z12 I2 ……… (1)

V2 = z21 I1 + z22 I2 ………. (2)

And, I1 = g11 V1 + g12 I2 ………. (3)

V2 = g21 V1 + G22 I2 ……….. (4)

Now, (3) and (4) can be rewritten as,

V1 = \(\frac{I_1}{g_{11}}– \frac{g_{12}}{g_{11}} I_2\) ………… (5)

And V2 = \(\left(g_{22} – \frac{g_{21} g_{12}}{g_{11}}\right) I_2 – \frac{g_{21} I_1}{g_{11}}\)……….. (6)

∴ Comparing (1), (2) and (5), (6), we get,

z11 = \(\frac{1}{g_{11}} \)

z12 = – \(\frac{g_{12}}{g_{11}} \)

z21 = – \(\frac{g_{21}}{g_{11}} \)

z22 = \(\left(g_{22} – \frac{g_{21} g_{12}}{g_{11}}\right)\).

Right choice is (a) z12 = \(\frac{1}{g_{11}} \)

Explanation: We know that, V1 = z11 I1 + z12 I2 ……… (1)

V2 = Z21 I1 + z22 I2 ………. (2)

And, I1 = g11 V1 + g12 I2 ………. (3)

V2 = g21 V1 + g22 I2 ……….. (4)

Now, (3) and (4) can be rewritten as,

V1 = \(\frac{I_1}{g_{11}}– \frac{g_{12}}{g_{11}} I_2\) ………… (5)

And V2 = \(\left(g_{22} – \frac{g_{21} g_{12}}{g_{11}}\right) I_2 – \frac{g_{21} I_1}{g_{11}}\)……….. (6)

∴ Comparing (1), (2) and (5), (6), we get,

z11 = \(\frac{1}{g_{11}} \)

z12 = – \(\frac{g_{12}}{g_{11}} \)

z21 = – \(\frac{g_{21}}{g_{11}} \)

z22 = \(\left(g_{22} – \frac{g_{21} g_{12}}{g_{11}}\right)\).

Right answer is (a) 0.56 mA

Easy explanation: INITIAL current immediately after CHARGING is GIVEN by,

\(\frac{V}{R} = \frac{5+3}{10000}\)

= 0.8 mA

Now, i = i0e^-t/RC

= 0.8 mA x \(e^{\frac{t}{10k X 10 X 10^{-6}}}\)

= 0.8 X 10^-3 X \(e^{\frac{35 X 10^{-3}}{10^{-1}}}\)

= 0.56 mA.

Correct ANSWER is (c) 5 TRIANGULAR

Easy EXPLANATION: We KNOW that,

Q = CV

Or, 0.1 X 50 = 5

And it is a triangular pulse.