Question

Hello guys..


Answer this plz​

Answer 1

R=100Ω,C=10μF,L=0.1803H

for heating by 10

∘

C, we need =10

∘

C×2J/C

=20J of heat

z=

100

2

+(2π×750×1803−

2π×750×10

−5

1

)

2

z=834Ω,Irms=

834

20

=0.024A

Paug=VrmsIrmsCosθ

=20×024×

834

100

=2(0.029)J/S=0.058J/s

Total energy=20=(t)Paug

20=t×(0.029)×2

348sec=t

all energy of the circuit is dissipated into resistor

one anerage there is no energy flow into capacitor or inductor

Answer 2

$\Huge\bf\maltese{\underline{\green{Answer°᭄}}}\maltese$

$\implies$$\large\bf{\underline{\red{VERIFIED✔}}}$

$R=100Ω,C=10μF,L=0.1803H \\ for \: heating \: by  \: 10∘C,\\ we need =10∘C×2J/C \\ =20J of heat \\ z= \sqrt{ {100}^{2} +1002+(2π×750×1803− \frac{1}{2π×750×10^{ - 5}} } \\z=834Ω,Irms=83420=0.024A \\Paug=VrmsIrmsCosθ \\ =20×024×834100 \\=2(0.029)J/S=0.058J/s \\ Total energy=20=(t)Paug \\ 20=t×(0.029)×2 \\ 348sec=t$

$\boxed{I \:Hope\: it's \:Helpful}$

${\sf{\bf{\blue{@ℐᴛz ᴅɪɴᴜ࿐}}}}$

• Please Verify It :)