Hola! ♥️ pls solve it!!

1.	Hola! ♥️ pls solve it!!
Answer» Explanation: The angular frequency of the LCR circuit is: $w(omega) = \frac{1}{ \sqrt{lc} }$ here: L=1.5 H ; C=35 micro farad ; R= 20 OHM putting the value in the above formulae, we get: $= > \frac{1}{ \sqrt{(1.5) \times (35 \times {10}^{ - 6}) } }$ => 138 Hz *When* an AC *voltage* of *Angular* *frequency* w(*omega) =138* Hz is *applied* to *the* *circuit, then* w(*omega) L= $138 \times 1.5 = <klux>207</klux>$ =>So inductive* RESISTANCE will be= 207 ohm Also, $capacitive \: reactance = \frac{1}{w(omega) \times c}$ $\frac{1}{138 \times (35 \times {10}^{ - 6}) } = 207 \: ohm$ $impendence (Z) = \sqrt{ {r}^{2} + {(wl \: - \frac{1}{wc} )}^{2} }$ This the condition of resonance. The rms resonant current in the circuit is: => I=V/Z i.e. $\frac{200}{20} = 10$ So, *resonant* *current* *will* be: 10 A *Note: here* *The* *power* is *transferred* to *the* *resistor* *only* *therefore* : *Average* *power= (II) R => (10A* * *10A) 20 *ohm* = *2000* W *Hence, the* *average* *power* *transferred* is *2000* W. HOPE IT HELPS YOU

Answer»

Explanation:

The angular frequency of the LCR circuit is:

$w(omega) = \frac{1}{ \sqrt{lc} }$

here:

L=1.5 H ; C=35 micro farad ; R= 20 OHM

putting the value in the above formulae, we get:

$= > \frac{1}{ \sqrt{(1.5) \times (35 \times {10}^{ - 6}) } }$

=> 138 Hz

When an AC voltage of Angular frequency w(omega) =138 Hz is applied to the circuit, then

w(omega) L=

$138 \times 1.5 = <klux>207</klux>$

=>So inductive RESISTANCE will be= 207 ohm

Also,

$capacitive \: reactance = \frac{1}{w(omega) \times c}$

$\frac{1}{138 \times (35 \times {10}^{ - 6}) } = 207 \: ohm$

$impendence (Z) = \sqrt{ {r}^{2} + {(wl \: - \frac{1}{wc} )}^{2} }$

This the condition of resonance. The rms resonant current in the circuit is:

=> I=V/Z

i.e.

$\frac{200}{20} = 10$

So, resonant current will be: 10 A

Note: here The power is transferred to the resistor only

therefore :

Average power= (I*I) *R

=> (10A * 10A)* 20 ohm = 2000 W

HOPE IT HELPS YOU