In a series LCR circuit the voltage across resistance, capacitance end inductance is 10 V each.
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Answer:
If the capacitance is made to be short, the voltage in the inductance will be 10/√2. The current in the circuit is 'I'. The impedance in the resistor is 'R'. The impedance across the inductance is wL = X_L. The impedance across capacitance is 1/wc = X_C. The total impedance of L and C is X = X_C - X_L. We know here that X_L = R = X_C as the voltages are the same. Hence, complete impedance in the circuit = √[ R² + X² ]. The voltage in the Inductor is behind the current that flows through the resistor. The voltage in the capacitor leads the current by 'R'. When V_R = V_L = V_C, i.e., the voltages across resistor, inductor and capacitors are all the same, then I * X_L = I * X_C, but in opposite phase. Hence, they cancel out. The total impedance is X_L - X_C = 0. So, the total voltage across all the three elements is 10 V. Therefore, the voltage that is applied in all these three elements is 10 V. If the capacitance gets shorted now, then only the inductor and resistor remains in the circuit. Hence, Z = √[R² + X_L²] = √2 R. An external source of 10 V is there. Thus, the current I = 10 /(√2R) amp. Therefore, the potential difference in L : I X_L = I R = 10/√2 V