JEE MAIN - Physics (2011 - No. 6)
A resistor $$'R'$$ and $$2\mu F$$ 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. $$\left( {{{\log }_{10}}2.5 = 0.4} \right)$$
$$1.7 \times {10^5}\,\Omega $$
$$2.7 \times {10^6}\,\Omega $$
$$3.3 \times {10^7}\,\Omega $$
$$1.3 \times {10^4}\,\Omega $$
Explanation
We have, $$V = {V_0}\left( {1 - {e^{ - t/RC}}} \right)$$
$$ \Rightarrow 120 - 200\left( {1 - {e^{ - t/RC}}} \right)$$
$$ \Rightarrow t = RC\,in\,\left( {2.5} \right)$$
$$ \Rightarrow R = 2.71 \times {10^6}\Omega $$
$$ \Rightarrow 120 - 200\left( {1 - {e^{ - t/RC}}} \right)$$
$$ \Rightarrow t = RC\,in\,\left( {2.5} \right)$$
$$ \Rightarrow R = 2.71 \times {10^6}\Omega $$
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