JEE Main · 2011 · 10 Apr 2011 · Offline · Current Electricity

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)

(A) $1.7 \times 1 0^{5} Ω$
(B) $2.7 \times 1 0^{6} Ω$ ✓ Correct
(C) $3.3 \times 1 0^{7} Ω$
(D) $1.3 \times 1 0^{4} Ω$

Answer: (B)

Solution

While the capacitor charges, the voltage across it rises as

V_{C} = E (1 - e^{- t / R C})

with

E = 200 V

. The neon bulb strikes when

V_{C} = 120 V

t = 5 s

, so

1 - e^{- 5/ R C} = \frac{120}{200} = 0.6

, giving

e^{- 5/ R C} = 0.4 = \frac{1}{2.5}

. Taking logarithms,

\frac{5}{R C} = ln 2.5 = 2.303 lo g_{10} 2.5 = 2.303 \times 0.4 = 0.921

. With

C = 2 \times 1 0^{- 6} F

this gives

R = \frac{5}{0.921 \times 2 \times 1 0 ^{- 6}} \approx 2.7 \times 1 0^{6} Ω

JEE Main · 2011 · 10 Apr 2011 · Offline · Single correct option

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JEE Main · 2011 · 10 Apr 2011 · Offline · Current Electricity

Solution

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