An inductor and a capacitor having reactances of 40 and 65...

An inductor and a capacitor having reactances of 40Ω and 65Ω respectively, are connected in series to a resistor of 20Ω to an AC source of 220V, 50Hz. Calculate the (i) Inductance of the inductor (ii) Capacitance of the capacitor (iii) Impedance of the circuit (iv) Circuit current (v) Voltage drop across the resistor, inductor, and capacitor.

Transcript

00:01 So here it is given that an inductor and capacitor having the resistance of 40 ohm and 65 ohm respectively, connected in series with the resistor of 20 oom to an ac source of 220 volt supply with 50 hertz frequency.

00:22 So we have to calculate the parts from 1 to 4.

00:26 Okay, so generally the resistance of inductor is denoted by xl while resistant for capacitor is given by xc and resistant for resistor let's suppose is capital r and voltage is v while this frequency to be as f so first we will try to find the omega here which is nothing but is equals to the two -par -f so it will be equals to 2 .2 50 comes out 100 pi now if we substitute the value pie here that is 100 multiplied by 3 .14 so it will be equals to 314 radium per second now from the first part we have to calculate the inductance of this circuit okay so simply we know that the inductive reactants or resistance of inductor xl is given by xl is equal to omega and 12 from here l will be quashed to xl upon omega so it will be quashed to xl is 40 ohm divided by omega is 314 so it comes out 0 .1274 and the unit will be henry now in this second part we have to calculate the capacitance of the capacitor.

02:00 So generally we know that the capacity reactants or capacitor resistance is given by xx is equal to 1 upon omega c.

02:10 So from here we can write c will be equals to 1 upon omega into xc so it will be equals to 1 upon 314 multiply by xc is 65 oam so after calculating c comes out 0 .4899 into 10 rest to minus 6 ferret while we can approximate it to 0 .5 macro ferret.

02:46 Now in the third part we have to calculate the impedance of the circuit.

02:53 Okay, according to question we have given l, c and r are connected in series.

02:59 So for series lcr circuit, the impedance is denoted by z and given by the formula of under root of r squared plus xl minus xe that whole square.

03:13 So it will be equal to under root of r is 20.

03:18 So it's a square plus xl is 40 minus 65 that whole square.

03:24 So after calculating this, it comes out under root of 400.

03:30 Plus 40 minus 65, 15, 15 square will be 2 to 5.

03:35 So it will be quashed to under root of 625.

03:37 625 is a square of 25.

03:39 So it will be quast to 25.

03:43 Now, in the fourth part, we have to calculate this circuit current.

03:49 So generally, circuit current i is given by i -quest -2...