SSC JE Electrical Engineering 24 Jan 2018 Morning Paper

Question : 149 of 199

Marks: +1, -0

What will be the value of the unknown inductance

L_X

(in mH) and the effective resistance

R_X

(in Ω) respectively for the circuit given below, when no current flows through the galvanometer (G)?

Solution:

Now under balance bridge condition

Z_xZ_4=Z_2Z_3

∴(R_x+jωL_x)[{R_4×1/{jωc_4}}/{R_4+1/{jωc_4}}]

=R_2R_3

Equation real and imaginary part
we get

R_x={R_2R_3}/R_4

and

L_x=R_2R_3C_4

Now

R_x={R_2×R_3}/R_4={20×40}/30

= 80/3\Ω

and

L_x = R_2 × R_3 × C_4

= 20 × 40 × 40 × 10^{–6}

∴L_x=32\mH

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