A resistor R and 2 μF capacitor in series 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. (log₁₀ 2.5 = 0.4)

1.7 x 10⁵ Ω

2.7 x 10⁶ Ω

3.3 x 10⁷ Ω

1.3 x 10⁴ Ω

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Solution

2.7 x 10⁶ Ω

straight v space equals space 200 left parenthesis 1 minus straight e to the power of negative straight t divided by straight t end exponent right parenthesis 120 space equals space 200 space left parenthesis 1 minus straight e to the power of negative straight t divided by straight T end exponent right parenthesis straight e to the power of negative straight t divided by straight T end exponent space equals space fraction numerator 200 minus 120 over denominator 200 end fraction space equals space 80 over 200 straight t divided by straight T space equals space log space left parenthesis 2.5 right parenthesis space equals space 0.4 5 space equals space left parenthesis 0.4 right parenthesis space straight x space space straight x space 2 space straight x space 10 to the power of negative 6 end exponent rightwards double arrow space straight R space equals space fraction numerator 5 over denominator left parenthesis 0.4 right parenthesis space straight x space 2 space straight x space 10 to the power of negative 6 end exponent end fraction space straight R space equals space 2.7 space straight x space 10 to the power of 6

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