MGVCL Exam Paper (30-07-2021 Shift 3) The current in the coil of a large electromagnet falls from 6 A to 2 A in 10 ms. The induced emf across the coil is 100 V. Find the self-inductance of the coil. 0.25 H 0.5 H 1.25 H 0.75 H 0.25 H 0.5 H 1.25 H 0.75 H ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Equation of self inductance:V = Ls*(di/dt)Ls = V/(di/dt)= 100/((6 - 2)/(0.01))= 0.250 H= 250 mH
MGVCL Exam Paper (30-07-2021 Shift 3) A conductor having surface density is embedded in a dielectric medium of permittivity. The electric field in the medium is E. If it is known that the pressure p on the conductor surface is equal to the electric energy density in the medium, then p (in SI unit)is given by σ²/(2πε) σ²/(2ε) σ/(4πε) σ²/(4π) σ²/(2πε) σ²/(2ε) σ/(4πε) σ²/(4π) ANSWER EXPLANATION DOWNLOAD EXAMIANS APP As the point is near to the conductor will act as infinite sheetElectric field (E) is given by:E = σ²/(2ϵ)
MGVCL Exam Paper (30-07-2021 Shift 3) Replace the braketed phrase grammatically and conceptually with the help of the given opation. If the given sentence is correct then select the option ’The given sentence is correct’.The corono virus continued to spread, with the development of infections advancing quickly continue to spread, with the development of infections The given sentence is correct continued for spread, with the development of infections continued to spreading, with the development of infections continue to spread, with the development of infections The given sentence is correct continued for spread, with the development of infections continued to spreading, with the development of infections ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) In a series RLC circuit, the supply voltage is 230 V at 50 Hz. The resonant current is 2 A at the resonant frequency of 50 Hz. Under the resonant condition, the voltage across the capacitor is measured to be equal to 460 V. What are the values of L and C? 0.73 H and 13.85 pF 0.73 mH and 13.85 pF 0.53 H and 15.83 pF 0.53 mH and 15.83 pF 0.73 H and 13.85 pF 0.73 mH and 13.85 pF 0.53 H and 15.83 pF 0.53 mH and 15.83 pF ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Vc = I*Xc460 = 2*(1/(2*ᴨ*50*C))C = 2/(460*2*50*ᴨ)= 1.385*10⁻⁵ F= 13.85 FV_L = I*X_LL = V/(I*2*ᴨ*f)= 460/(2*ᴨ*50*2)= 0.7345 H
MGVCL Exam Paper (30-07-2021 Shift 3) The current flowing in to a balanced delta connected load through line 'a' is 10 A when the conductor of line ’b’ is open. With the current in line 'a' as reference, compute the symmetrical components of the line currents. Assume the phase sequence of ’abc’. Ia₀ = 0 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 10 A.Ia₁ = (-5 + j2.44) A.Ia₂ = (5 - j2.44) A. Ia₀ = 10 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 0 A.Ia₁ = (-5 + j2.88) A.Ia₂ = (5 - j2.88) A. Ia₀ = 0 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 10 A.Ia₁ = (-5 + j2.44) A.Ia₂ = (5 - j2.44) A. Ia₀ = 10 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 0 A.Ia₁ = (-5 + j2.88) A.Ia₂ = (5 - j2.88) A. ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For the given question,Ib = 0Ia = -IcCalculation:Ia1 = 1/3*(Ia + a*Ib + a²*Ic)= 1/3*(10 + 10∟-120°)= 5 + j*2.88 AIa2 = 1/3*(Ia + a²*Ib + a*Ic)= (1/3)*(10 + 10∟120°)= 5 - j*2.88 AIao = (1/3)*(Ia + Ib + Ic)= (Ia - Ia)= 0
MGVCL Exam Paper (30-07-2021 Shift 3) મેં મીઠાઈ બનાવી'. વાક્યને કેવળ ક્રિયાપદની પ્રેરક રચના કઈ છે? મેં મીઠાઈ બનાવી લીધી. મેં મીઠાઈ બનાવડાવી. હું મીઠાઈ બનાવવા લાગી. મીઠાઈ તો હું જ બનવું ને! મેં મીઠાઈ બનાવી લીધી. મેં મીઠાઈ બનાવડાવી. હું મીઠાઈ બનાવવા લાગી. મીઠાઈ તો હું જ બનવું ને! ANSWER DOWNLOAD EXAMIANS APP
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