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JEE Mains Previous Paper 1 (Held On: 08 Apr 2019 Shift 2)

Option 4 : RC circuit with R = 1 kΩ and C = 10 μF

JEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)

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90 Questions
360 Marks
180 Mins

**Concept:**

**Capacitive Reactance X _{c}:**

Capacitive reactance (X_{c}) is large at low frequencies and small at high frequencies. For steady DC which is zero frequency (f = 0Hz), X_{c} is infinite (total opposition), which means that capacitors pass AC but block DC.

At resonance, the reactance will be same for both capacitances.

X_{C }= R

\(\because {{\rm{X}}_{\rm{C}}} = \frac{1}{{2\pi fC}}\)

Where

X_{c} = reactance in ohms (ohm)

f = frequency in hertz (Hz)

C = capacitance in farads (F)

**Calculation:**

We know that, ω = 2πf

\( \Rightarrow R = \frac{1}{{\omega C}}\)

\( \Rightarrow RC = \frac{1}{\omega }\)

ω = 100 rad/s

\( \Rightarrow RC = \frac{1}{{100}}\)

∴ RC = 10^{-2}

The RC circuit with R = 1 kΩ and C = 10 μF exhibits this wave.

On substituting these values,

⇒ RC = 1 × 10