A resistor 'R' and 2 capacitor in series is connected through a switch to 200 V direct supply. Across the capacitor is a neon bulb that lights up at 120 V. Calculate the value of R to make the bulb light up 5 s after the switch has been closed. (log ₁₀2.5=0.4) (a) 1.7× 10⁵ Ω (b) 2.7× 10⁶ Ω (c) 3.3× 10⁷ Ω (d) 1.3× 10⁴ Ω

Answer: (b). Charging: V = V₀(1 - e-t/RC) 120 = 200(1 - e-t/RC) ⇒ e-t/RC = 0.4 ⇒ t/RC = ln 2.5 = 2.303 × 0.4 = 0.921 R = (t)/(0.921 C) = 50.921 × 2×10⁻⁶ R ≈ 2.7×10⁶ Ω So the answer is (b) 2.7× 10⁶ Ω.

A resistor 'R' and 2 capacitor in series is connected through a… | JEE Main 2011 (AIEEE)

Q43JEE Main 2011 · AIEEEDifficult

Capacitors in DC Circuits & RC

A resistor $'R'$ and $2\ \mu\text{F}$ capacitor in series is connected through a switch to $200\text{ V}$ direct supply. Across the capacitor is a neon bulb that lights up at $120\text{ V}.$ Calculate the value of $R$ to make the bulb light up $5\text{ s}$ after the switch has been closed. $(\log_{10}2.5=0.4)$

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Answer: (b)

Charging: $V = V_0\left(1 - e^{-t/RC}\right)$
$120 = 200\left(1 - e^{-t/RC}\right) \Rightarrow e^{-t/RC} = 0.4 \Rightarrow \frac{t}{RC} = \ln 2.5 = 2.303 \times 0.4 = 0.921$
$R = \frac{t}{0.921\,C} = \frac{5}{0.921 \times 2\times10^{-6}}$
$R \approx 2.7\times10^{6}\ \Omega$
So the answer is (b) $2.7\times 10^{6}\ \Omega$ .

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