00:01
This is the given problem here.
00:10
This is the given problem here.
00:14
This is a series rlc circuit.
00:18
Resistance of 1 .8 kilo -oam is in series with a inductor of inductive reactance 2 kilo -oom and a capacitor of capacitive reactance 0 .6 kilo -oom.
00:29
They are in series and they are connected to a voltage source of of the value e equal to 6 sine 314 t plus 60 degree.
00:42
Right? so this is a series rlc circuit.
00:46
In first case, we are supposed to find out the total impedance of the circuit.
00:53
Now, as we know, the total impedance of a series rlc circuit, as per the standard formula is ri, say, plus xl minus xc into j.
01:04
This is in the rectangular form.
01:08
Now, if you want to find out the magnitude of the impedance, then it is equal to r equal to under root capital r squared, say plus xl minus xx, whole square.
01:22
We have all the values given in the diagram.
01:26
We just put the value.
01:27
So, impedance of the circuit will have the magnitude 2 .28 into 10 to 3 all.
01:35
Now what will be the face angle of the impedance from standard formula, phase angle is equal to 10 inverse xl minus xc upon r.
01:44
We have all the values.
01:46
We simply put the values, we will get the face angle 37 .84 degree.
01:51
So this is 2 .28 into 10 raised to 3 oom.
01:56
Angle 37 .84 degree is the total impedance of the circuit in the phaser form or in the polar form.
02:06
Right fine in part b we are asking us to find out the impedance triangle now here in case of the means resistance the voltage and current are in phase so we are marking this like this are side and in case of inductor the voltage leads the current by by 90 degree so we are marking it it upward, means plus 90 degree, and in case of a capacitor is lags by 90 degree, we are marking it minus 90 degree.
02:43
If we do the vector addition of xl and xc, normally xc is much less than xl, so the difference will be xl minus xe, right? and this is r, and using the vector addition formula, we can find out the means magnitude of total impedance, this is a right angle triangle, means one side is base is r and the perpendicular is xl minus xc.
03:12
So its hypotenous will be under root r squared plus xl minus xc.
03:19
And this angle 5, we have already calculated, is 37 .84, means degree.
03:25
So this triangle is nothing but the impedance triangle, which shows on all the, on the diagonal, it shows the total impede.
03:33
On the perpendicular, it shows the reactive part, i mean capacitive and the inductive part, and the base is the resistive part and the angle, phi is the face angle.
03:47
We can choose any angle, any scale like 2 cm is equal to 1 into 10 to 3 om and draw the actual impedance triangle.
03:58
Now in part c, they are saying phi inter means current and the voltage is in the phaser form.
04:07
Now let us first write this is a given voltage equal to e0, sine, means omega t plus 5.
04:16
This is a standard equation.
04:18
We compare the given voltage with this equation.
04:21
We can see the face angle for the voltage is 60 degree.
04:27
And now what is the magnitude of the voltage? for the magnitude we take root mean square values.
04:33
In general we take root mean square value which is peak value divided by root 2, peak value is 6, so it is 6 by root 2, it means 4 .24 volt.
04:43
So the applied voltage, we can write it in as a polar form, its root mean square value and the angle.
04:53
Done.
04:54
And what is the total current? now total current is, say, from home's law, are equal to v by r.
05:00
So this is the applied voltage and the total impedance of the circuit.
05:07
So this is already no and total impedance we have means already calculated in the previous parts.
05:15
So when we do the calculation, the 60 degrees in the numerator and 37 .84 degree in the denominator.
05:24
So when we will solve it, this will be 60 minus 37 .84 and this part will be solved as normally.
05:32
So this will be 1 .86 and angle will be 22 .16, say because 60 minus 37 .84 is 22 .16.
05:46
These are all the root minus square values.
05:48
And also the voltage drop at the resistance root minus square value v equal to ir is the total current root minus square value into multiplied by the value of the resistance.
06:00
Now resistance has no angle because here voltage and current are in phase.
06:06
So we are not writing any angle with the resistance.
06:10
They have all the values means given.
06:12
So we simply multiply them.
06:14
This is a voltage drop at the resistance.
06:18
Now voltage drop at the inductor.
06:20
It will be the total current multiplied by the inductive reactance.
06:25
Now the inductive reactance is given as 2 kilo -on.
06:29
That's true...