Question

A series RC combination is connected to an AC voltage of angular frequency 500rad/s, if the impedance of the RC circuit is R $\sqrt{1.25}$, the time constant of the circuit is?

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Answer 1

$\huge\purple{Solution:-}$

ɢɪᴠᴇɴ :

ω = 500 ʀᴀᴅɪᴀɴ/s

ʟᴇᴛ ᴛʜᴇ ᴄᴀᴘᴀᴄɪᴛᴀɴᴄᴇ ᴏғ ᴛʜᴇ ᴄᴀᴘᴀᴄɪᴛᴏʀ ʙᴇ ᴄ.

ᴛʜᴜs ʀᴇsɪsᴛᴀɴᴄᴇ ᴏғ ᴄᴀᴘᴀᴄɪᴛᴏʀ xᴄ =

1/ωᴄ = 1/500ᴄ

• ɪᴍᴘᴇᴅᴀɴᴄᴇ ᴏғ ᴛʜᴇ ᴄɪʀᴄᴜɪᴛ ᴢ = ʀ√1.25

ᴜsɪɴɢ ᴢ² = ʀ² + x²ᴄ

∴ 1.25ʀ² = ʀ² + 1/(500)²ᴄ²

ᴏʀ 0.25ʀ² = 1/0.25 × 10^6ᴄ²

⇒ʀ²ᴄ² = 10^-6/(0.25)²

ᴡᴇ ɢᴇᴛ ᴛɪᴍᴇ ᴄᴏɴsᴛᴀɴᴛ ᴏғ ᴛʜᴇ ᴄɪʀᴄᴜɪᴛ ʀᴄ= 10^-3/0.25

= 0.004s =4 ᴍs

ʜᴏᴘᴇ ɪᴛ ʜᴇʟᴘs! ᴛʜᴀɴᴋ ʏᴏᴜ!

Answer 2

${\large{\underbrace{\textsf{\textbf{\color{red}{⟹ }{\color{green}{\:\:ANSWER\: :-}}}}}}}$

Given : ω=500 radian/s

Let the capacitance of the capacitor be C.

Thus resistance of capacitor XC

$→ = ωC1 = 500C1$

$\sf Impedance\; of\; the\; circuit Z=R1.25$

$→Using Z2 =R 2 +X C2$

$∴1.25R2 =R 2 + (500) 2 C 21$

$\tt Or0.25R 2=25×10 6 C 21⟹R 2 C 2 = (0.25) 210−6$

We get time constant of the circuit $\tt →RC=0.2510 −3 =0.004 s =4 ms$