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DSSSB JE EE 2019 Official Paper (Held on 25 Oct 2019)

Option 3 : 50%

**Before Fault:**

V_{L} = Line voltage

I_{R} , I_{Y} , I_{B} are line currents

R_{1} is the resistance of the load

V_{Ph} = Phase Voltage = V_{L}/√3

\(I_R = I_Y=I_B={V_L \over \sqrt3 × R_1}\)

Power loss in load = 3 × I_{R}^{2} × R_{1}

\(=3 \times ({V_L \over \sqrt3 × R_1})^2\times R_1= {V_L^2 \over R_1}\)**.......(1)**

**After fault and one line removed:**

VL = Line voltage

IR , IY , IB are line currents

IB = 0

R2 is the resistance of the load

\(I_R = -I_Y={V_L \over 2 × R_2}\)

Power loss in load = 2 × IR2 × R2

\(=2 \times ({V_L \over 2 × R_2})^2\times R_2= {V_L^2 \over 2R_2}\)**.....(2)**

from given data power losses are equal. So equations (1) and (2) are equal

\( {V_L^2 \over R_1}={{V_L^2 \over 2R_2}}\)

\( { R_2}={R_1\over2}\)

% change is given by

\({{R_1\over2}-{R_1} \over {R_1}}\times 100 = -50\)

-ve sign indicates decrease

decreased by **50%**