A resistor R and 2 μF capacitor in series is connected through a switch to 200 V direct supply. Across the capacitor is a neon bulb that lights up at - Physics MCQ FIITJEE AIEEE 2011 (Set Q) 2026 | ExamTest.live

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A resistor R and 2 μ F 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. (lo g_{10} 2.5 = 0.4)

Solution & Step-by-step Explanation

Voltage across capacitor in charging RC circuit: V_{c} = E (1 - e^{- t / RC}) 120 = 200 (1 - e^{- 5/ (R \times 2 \times 1 0^{- 6})}) 0.6 = 1 - e^{- t / τ} ⟹ e^{- t / τ} = 0.4 ⟹ e^{t / τ} = 2.5 Taking natural log: \frac{t}{τ} = ln (2.5) = 2.303 lo g_{10} (2.5) \frac{5}{R \times 2 \times 1 0 ^{- 6}} = 2.303 \times 0.4 \approx 0.92 R = \frac{5}{2 \times 1 0 ^{- 6} \times 0.92} \approx 2.7 \times 1 0^{6} Ω

Try it yourself before checking the explanation above.

A resistor R and 2 μ F 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. (lo g_{10} 2.5 = 0.4)

1.7 \times 1 0^{5} Ω

2.7 \times 1 0^{6} Ω

3.3 \times 1 0^{7} Ω

1.3 \times 1 0^{4} Ω

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