SC JE Tier-II Electrical – Solutions

Q1 – Induction Motor (Rotor Current, Air-Gap Power, Rotor Copper Loss)

Given:

• $V_L = 400 \, V$VL​=400V, star connection → $V_{ph} = \frac{400}{\sqrt{3}} = 230.94 \, V$Vph​=3​400​=230.94V
• $R_2 = 0.05\ \Omega$R2​=0.05 Ω, $X_2 = 0.15\ \Omega$X2​=0.15 Ω
• Slip $s = 0.02$s=0.02

Step 1 – Rotor impedance referred to stator:$$Z_2′ = \frac{R_2}{s} + j X_2 = \frac{0.05}{0.02} + j 0.15 = 2.5 + j0.15\ \Omega$$Z2′​=sR2​​+jX2​=0.020.05​+j0.15=2.5+j0.15 Ω

Step 2 – Rotor current per phase:$$I_2 = \frac{V_{ph}}{Z_2′} = \frac{230.94}{\sqrt{2.5^2 + 0.15^2}} = \frac{230.94}{2.504} \approx 92.2\ A$$I2​=Z2′​Vph​​=2.52+0.152​230.94​=2.504230.94​≈92.2 A

Step 3 – Air-gap power PagP_{ag}Pag​:$$P_{ag} = 3 I_2^2 \frac{R_2}{s} = 3 (92.2)^2 \times 2.5 \approx 63.6\ kW$$Pag​=3I22​sR2​​=3(92.2)2×2.5≈63.6 kW

Step 4 – Rotor copper loss:$$P_{rotor} = 3 I_2^2 R_2 = 3 (92.2)^2 \times 0.05 \approx 1.27\ kW$$Protor​=3I22​R2​=3(92.2)2×0.05≈1.27 kW

✅ Answer Q1:

• Rotor current = 92.2 A
• Air-gap power ≈ 63.6 kW
• Rotor copper loss ≈ 1.27 kW

Q2 – Transformer (Efficiency & Voltage Regulation)

Given:

• Transformer rating: 50 kVA, 3-phase, delta 400 V
• Resistances: R1 = 0.5 Ω, R2 = 0.02 Ω
• Reactances: X1 = 2 Ω, X2 = 0.05 Ω

Step 1 – Equivalent impedance per phase:$$Z_{eq} = R_1 + R_2 + j(X_1 + X_2) = 0.52 + j 2.05\ \Omega$$Zeq​=R1​+R2​+j(X1​+X2​)=0.52+j2.05 Ω

Step 2 – Full-load current per phase:$$I_{FL} = \frac{V_{ph}}{Z_{load} + Z_{eq}} \approx \frac{230.94}{\sqrt{0.52^2 + 2.05^2}} \approx 110\ A$$IFL​=Zload​+Zeq​Vph​​≈0.522+2.052​230.94​≈110 A

Step 3 – Full-load copper loss:$$P_{cu} = 3 I_{FL}^2 (R_1 + R_2) = 3 \cdot (110)^2 \cdot 0.52 \approx 18.9\ kW$$Pcu​=3IFL2​(R1​+R2​)=3⋅(110)2⋅0.52≈18.9 kW

Step 4 – Full-load efficiency:$$\eta = \frac{S \cdot pf}{S \cdot pf + P_{cu} + P_{core}} \approx \frac{50 \cdot 0.8}{50 \cdot 0.8 + 18.9 + 1} \approx 68\%$$η=S⋅pf+Pcu​+Pcore​S⋅pf​≈50⋅0.8+18.9+150⋅0.8​≈68%

(Core loss assumed 1 kW)

Step 5 – Voltage Regulation: VR = \frac{|V_{NL}| – |V_{FL}|}{|V_{FL}|} \times 100 \approx 5–6\% \] (Typical) ✅ **Answer Q2:** – Efficiency ≈ 68% – Voltage Regulation ≈ 5–6% — ## **Q3 – Transmission Line (Sending/Receiving Voltage, Losses)** **Given:** – R = 0.1 Ω/km, X = 0.2 Ω/km, L = 50 km → Z = (5 + j10) Ω per phase – Load = 10 MW, 0.8 pf lagging, V_sending = 132 kV **Step 1 – Receiving end current:** \[ I_R = \frac{P + jQ}{\sqrt{3} V_R} = \frac{10 \times 10^6}{\sqrt{3} \cdot 132\times10^3 \cdot 0.8} \approx 54.7 A

Step 2 – Sending end voltage:$$V_S = V_R + I_R Z_{line} \approx 132\text{ kV } + (54.7 \cdot (5 + j10))$$VS​=VR​+IR​Zline​≈132 kV +(54.7⋅(5+j10)) $$V_S \approx 132 + (273.5 + j547) V \approx 132.58 kV, \angle 63°$$VS​≈132+(273.5+j547)V≈132.58kV,∠63°

Step 3 – Line losses:$$P_{loss} = 3 I_R^2 R = 3 (54.7)^2 \cdot 5 \approx 44.9\ kW$$Ploss​=3IR2​R=3(54.7)2⋅5≈44.9 kW

✅ Answer Q3:

• Receiving current ≈ 54.7 A
• Sending voltage ≈ 132.58 kV ∠63°
• Line losses ≈ 44.9 kW

Q4 – Moving Coil Ammeter Conversion

Given: I_fs = 100 A, R_m = 0.1 Ω, desired full-scale = 500 A

Step 1 – Shunt resistor:$$I_s = I_{total} – I_m = 500 – 100 = 400 A$$Is​=Itotal​−Im​=500−100=400A $$V_m = I_m R_m = 100 \cdot 0.1 = 10 V$$Vm​=Im​Rm​=100⋅0.1=10V $$R_{sh} = \frac{V_m}{I_s} = \frac{10}{400} = 0.025\ \Omega$$Rsh​=Is​Vm​​=40010​=0.025 Ω

Step 2 – New meter voltage drop:$$V_{drop} = I_m R_m = 10 V \ (same)$$Vdrop​=Im​Rm​=10V (same)

✅ Answer Q4:

• Shunt resistor = 0.025 Ω
• Voltage drop = 10 V

Q5 – Full-Bridge Rectifier

Given: V_RMS = 230 V, R_load = 10 Ω

Step 1 – Peak voltage:$$V_{m} = \sqrt{2} \cdot 230 \approx 325.3\ V$$Vm​=2​⋅230≈325.3 V

Step 2 – Average load voltage (full-wave):$$V_{dc} = \frac{2 V_m}{\pi} \approx \frac{2 \cdot 325.3}{3.1416} \approx 207 V$$Vdc​=π2Vm​​≈3.14162⋅325.3​≈207V

Step 3 – Load current:$$I_{dc} = \frac{V_{dc}}{R} = \frac{207}{10} \approx 20.7 A$$Idc​=RVdc​​=10207​≈20.7A

Step 4 – Peak inverse voltage of diode:$$PIV = V_m = 325.3 V$$PIV=Vm​=325.3V

✅ Answer Q5:

• Average voltage = 207 V
• Load current ≈ 20.7 A
• PIV = 325.3 V

Q6 – Series RLC Circuit

Given: R = 20 Ω, L = 0.2 H, C = 50 μF, V = 230 V, f = 50 Hz

Step 1 – Reactances:$$X_L = 2 \pi f L = 2 \pi \cdot 50 \cdot 0.2 \approx 62.83\ \Omega$$XL​=2πfL=2π⋅50⋅0.2≈62.83 Ω $$X_C = \frac{1}{2 \pi f C} = \frac{1}{2 \pi \cdot 50 \cdot 50 \times 10^{-6}} \approx 63.66\ \Omega$$XC​=2πfC1​=2π⋅50⋅50×10−61​≈63.66 Ω

Step 2 – Total impedance:$$Z = \sqrt{R^2 + (X_L – X_C)^2} = \sqrt{20^2 + (62.83-63.66)^2} \approx \sqrt{400 + 0.69^2} \approx 20.01\ \Omega$$Z=R2+(XL​−XC​)2​=202+(62.83−63.66)2​≈400+0.692​≈20.01 Ω

Step 3 – Circuit current:$$I = \frac{V}{Z} = \frac{230}{20.01} \approx 11.5\ A$$I=ZV​=20.01230​≈11.5 A

Step 4 – Power factor:$$\cos \phi = \frac{R}{Z} = \frac{20}{20.01} \approx 0.9995 \ (\text{almost unity})$$cosϕ=ZR​=20.0120​≈0.9995 (almost unity)

Step 5 – Resonant frequency:$$f_r = \frac{1}{2 \pi \sqrt{LC}} = \frac{1}{2 \pi \sqrt{0.2 \cdot 50 \times 10^{-6}}} \approx 159 Hz$$fr​=2πLC​1​=2π0.2⋅50×10−6​1​≈159Hz

✅ Answer Q6:

Resonant frequency ≈ 159 Hz

Z ≈ 20 Ω

I ≈ 11.5 A

pf ≈ 0.9995 (unity)