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AP EAMCET201920 Apr 2019Morning ShiftPhysicsAlternating CurrentActual

When an inductor of inductance ( 6 H ), a capacitor of capacitance ( 50 F ) and resistor of resistance (R ) are connected in series with an AC supply of rms voltage (220 ~V ) and frequency (50 ~Hz ), the rms current through the circuit is (440 ~mA ). Match the inductive reactance, (X_L ) the capacitive reactance, (X_C ) the resistance (R ) and the impedance (Z ) of the circuit given in List-I with the corresponding v

Options

  1. A( array cc A & B & C & D (iv) & (ii) & (i) & (iii) array )
  2. B( array cc A & B & C & D (iv) & (iii) & (i) & (ii) array )
  3. C( array cc A & B & C & D (iv) & (i) & (ii) & (iii) array )
  4. D( array cc A & B & C & D (i) & (iv) & (iii) & (ii) array )

Correct answer

C. ( array cc A & B & C & D (iv) & (i) & (ii) & (iii) array )

Step-by-step solution

Given, inductance of inductor, (L= 6 H ), capacitance of capacitor, (C= 50 F ) supply voltage, (V_ rms =220 ~V ), supply frequency, (f=50 ~Hz ) and supply current, (I_ rms =440 ~mA ) Now, (A) inductive reactance, (X_L= L ) ( aligned X_L & =2 f L ( =2 f) & =2 50 6 =600 X_L & =600 aligned ) (B) Capacitance reactance, (X_C= 1 C = 1 2 f C ) ( aligned X_C & = 10^6 2 50 50 X_C & = 10^6 2500 2 X_C=200 aligned ) (C) Resistance, (R ) ( ) Impedance of LCR circuit, (Z^2=R^2+ (X_L-X_C )^2 ) Putting the given values, we get ( a

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