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In a series LCR circuit the voltage across resistance, capacitance and inductance is 10V each. If the capacitance is short circuited, the voltage across

the inductance will be



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The current flowing in the circuit is I.
Impedance across resistor = R.
Impedance across inductance = wL  = X_L
impedance across capacitance = 1/wc  = X_C
combined impedance of  L and C :  X = X_C - X_L
we are given X_L = R = X_C  as potentials are equal.

Total impedance in the  circuit =    √[ R² + X² ]

The voltage across Inductor lags behind the current through the resistor.  The voltage across the capacitor leads the current through  R.

When V_R = V_L = V_C = potentials across resistor, inductor and capacitors are all equal, then I *  X_L = I * X_C,  and they are in opposite phase.  So they cancel.  The collective impedance is X_L - X_C = 0.    The net voltage across the  3 elements is only 10 V.  So the voltage applied across these elements, is 10 V.

  Now if the capacitance is shorted, only inductor and resistor are there in the circuit.  So Z = √[R² + X_L²]  = √2 R.  There is an external voltage source of 10 V.
         So current I = 10 /(√2R)  amp.
   P.d. across L :  I X_L = I R = 10/√2  V


  The voltages across Resistance and voltage across inductor have a phase difference of  90 deg.  Hence, the net voltage =  √[ V_L² + V_R² ]  = 10 V
       => √2 V_R = 10 V
       => V_R = V_L = 10/√2  Volts

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