In series `L-C-R` circuit, voltge drop across resistance is `8V`, across inductor is `6V` and across capacitor is `12V`. Then,

In series LCR circuit voltage drop across resistance is 8V, across inductor is 6V and across capacitor is 12V. Then

In series LCR circuit voltage drop across resistance is 8V, across inductor is 6V and across capacitor is 12V. Then

If i is the current in L-C-R series AC circuit then, formula of, (i) Voltage across resistance. (ii) Voltage across inductor. (iii) Voltage across capacitor.

In a series L-C-R circuit the voltage across resistance , capacitance and inductance is 10 V each. If the capacitance is short circuited, the voltage across the inductance will be

In a series L-C-R circuit the voltage across resistance , capacitance and inductance is 10 V each. If the capacitance is short circuited, the voltage across the inductance will be

A 120 V, 60 Hz a.c. power is connected 800 Omega non-inductive resistance and unknown capacitance in series. The voltage drop across the resistance is found to be 102 V , then voltage drop across capacitor is

Consider a series LCR circuit connected to an AC supply of 220V. If voltage drop across resistance R is V_R, voltage drop across capacitor is V_c =2V_R and that across inudctor coil is V_L = 2V_R then choose correct alternative(s)

Consider a series LCR circuit connected to an AC supply of 220V. If voltage drop across resistance R is V_R, voltage drop across capacitor is V_c =2V_R and that across inudctor coil is V_L = 2V_R then choose correct alternative(s)