SSC JE Electrical Previous Year Question Paper 2018-SET1|SSC JE 2018

Extension of Ammeter Range

The current range of a DC moving coil ammeter is extended by connecting a shunt resistance Rs (low resistance) across the coil, the circuit as shown in Figure

I = Total current = 5A
I_m = full-scale deflection current of ammeter = ?
I_sh = shunt current
R_m = resistance of the ammeter = 25Ω
R_sh = shunt resistance = 75Ω

The full-scale deflection current I_m is given as

$\begin{array}{l}{I_m} = \dfrac{{I \times {R_{sh}}}}{{{R_m} + {R_{sh}}}}\\\\{I_m} = \dfrac{{5 \times 75}}{{(25 + 75)}}\\\\{I_m} = 3.75A\end{array}$

The average value of a sin wave for the complete cycle i.e 2π is given as

V_avg = 2⁄π V_peak

100 = 2⁄π V_peak

V_Peak = 157 V

Or

V_avg = 0.636 × V_Peak

V_Peak = 157

Now Peak to Peak Voltage is given as

V_PK − V_PK = 2 × V_Peak

V_PK − V_PK = 2 × 157 = 314 V

V_PK − V_PK = 314 V

If current flows for time t then the production of heat is given as

H = I²R.t

Where

I = current = 1 A

R = Resistance = 10 ohms

t = time = 5 sec

H = 1² × 10 × 5

H = 50 watt-sec or Joule

Tellegen’s theorem is one of the most powerful theorems in network theory. The physical interpretation of Te|legen‘s theorem is the conservation of power. As per the theorem, the sum of powers delivered to or absorbed by all branches of a given lumped network is equal to 0 i.e. the power delivered by the active elements of a network is completely absorbed by the passive elements at each instant of time.

Tellegen’s theorem depends on KCL and KVL but not on the type of the elements. Tellegen theorem can be applied to any network linear or non-linear, active or passive, time-variant or time-invariant.

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Ques.35. Calculate the phase angle between the voltage and current phasor of a series RL circuit having the resistance of 65 Ohms and Inductive reactance of 37.53 ohms when supplied by a frequency of 60 Hz.

1. 15°
2. 30°✓
3. 45°
4. 60°

Show Explanation

The phase angle θ of a series RL circuit is given as

θ = tan⁻¹(X_L ⁄ R)

Where

X_L = Inductive Reactance = 37.53

R = Resistance = 65

θ = tan⁻¹(37.53 ⁄ 65)

θ = tan⁻¹(0.57)

θ = 30°

Given

Inductance L = 0.01 mH = 0.01 × 10⁻³

Capacitance C = 0.01 mF = 0.01 × 10⁻³

The resonance frequency of an L-C tank circuit is given as

$\begin{array}{l}{f_{LC}} = \dfrac{1}{{2\pi \sqrt {LC} }}\\\\ = \dfrac{1}{{2\pi \sqrt {0.01 \times {{10}^{ – 3}} \times 0.01 \times {{10}^{ – 3}}} }}\\\\ = \dfrac{{{{10}^3}}}{{2\pi \times 0.01}}\\\\{f_{LC}} = 15.91kHz\end{array}$

Given

Inductance L = 1.8 H

Resistance R = 90 Ohms

Voltage V = 20 V

Time = 20 ms = 20 × 10⁻³

The current in an RL series circuit is given as

${I_{\left( t \right)}} = \dfrac{V}{R}\left( {1 – {e^{ – Rt/L}}} \right)$

The voltage drop across the inductor is

V_L = L(di/dt)

The Circuit is shown in the figure

$\begin{array}{l}{I_{\left( t \right)}} = \dfrac{V}{R}\left( {1 – {e^{ – Rt/L}}} \right)\\\\{V_L} = L \times \dfrac{{20}}{R}\left( {{e^{ – Rt/L}}} \right)\dfrac{R}{L}\\\\{V_L} = 20\left( {{e^{ – Rt/L}}} \right)\\\\{V_L} = 20\left( {{e^{ – \dfrac{{20 \times {{10}^{ – 3}} \times 90}}{{1.8}}}}} \right)\\\\{V_L} = 20 \times {e^{ – 1}}\\\\{V_L} = 20 \times 0.367\\\\{V_L} = 7.35V\end{array}$

Delta (Δ) to (Y) star conversion: We can convert the delta connection into its equivalent star connection with the following equations:

So, the equivalent of the impedance of each branch of the star connection is obtained by the multiplication of the impedances of the two delta branches that meet at its end divided by the sum of three delta impedances.

R_A, R_B, R_C = Product of an adjacent resistor ⁄  Sum of all resistor in delta

Given

R_AB = R_BC = R_CA = 36Ω

Therefore the delta to star conversion will be

$\begin{array}{l}{R_A} = \dfrac{{{R_{AB}} \times {R_{CA}}}}{{{R_{AB}} + {R_{BC}} + {R_{CA}}}}\\\\{R_A} = \dfrac{{36 \times 36}}{{36 + 36 + 36}}\\\\{R_A} = 12\Omega \\\\{R_B} = \dfrac{{{R_{BC}} \times {R_{AB}}}}{{{R_{AB}} + {R_{BC}} + {R_{CA}}}}\\\\{R_B} = \dfrac{{36 \times 36}}{{36 + 36 + 36}}\\\\{R_B} = 12\Omega \\\\{R_C} = \dfrac{{{R_{CA}} \times {R_{BC}}}}{{{R_{AB}} + {R_{BC}} + {R_{CA}}}}\\\\{R_C} = \dfrac{{36 \times 36}}{{36 + 36 + 36}}\\\\{R_C} = 12\Omega \end{array}$

So the value of per phase resistor in the equivalent star connection is 12 ohms.

Power consumption in an AC circuit is given as

P = VI cosφ

Given

Impedance Z = 16Ω

Current I = 4A

Power = 200 W

Power factor cosφ = ?

Voltage V = I × Z = 16 × 4 = 64 V

200  = 64 × 4 × cosφ

cosφ = 0.78

Power Factor = 0.78

When the load is balanced the Power consumed by the two wattmeters is

P = W1 + W2 = √3 V_L I_L cosφ

Where

V_Ph = Phase voltage = 220 V

Line voltage = √3 × V_ph = √3 × 200

I_Ph = Phase current = Line current I_L = 10 A

Power factor cosφ = 0.8

∴ Total Power consumed

P = √3 × √3 × 220 × 10 × .8

W₁ + W₂ = 5280 watt

Now

Sin²φ = 1 − cos²φ

Sin²φ = 1 − (0.8)²

Sinφ = 0.6

Now tanφ = Sinφ ⁄ Cosφ = 0.6 ⁄ 0.8 = 0.75

The power factor of the two wattmeters is given as

$\begin{array}{l}\tan \Phi = \sqrt 3 \left( {\dfrac{{{W_1} – {W_2}}}{{{W_1} + {W_2}}}} \right)\\\\0.75 = \sqrt 3 \left( {\dfrac{{{W_1} – {W_2}}}{{5280}}} \right)\\\\{W_1} – {W_2} = \dfrac{{3960}}{{\sqrt 3 }}\\\\{{\rm{W}}_{\rm{1}}}{\rm{ – }}{{\rm{W}}_{\rm{2}}}{\rm{ = 2286}}{\text{.3 watt}}\end{array}$