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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.

$\left ( \log_{10}2.5= 0.4 \right )$

Option 1)
$1.3 \times 10^{4}\Omega$

Option 2)
$1.7 \times 10^{5}\Omega$

Option 3)
$2.7 \times 10^{6}\Omega$

Option 4)
$3.3 \times 10^{7}\Omega$

Answers (1)

best_answer

As we learnt in

Current through the circuit

Self Inductance -

An emf is induced in the coil or the circuit which oppose the change that causes it. Which is also known back emf.

- wherein

$I=I_{o}.e^{-t/RC}$

$q=q_{o}.(1-e^{-t/RC})$

Potential difference across capacitor

$v=\frac{q}{c}=\frac{q_{o}}{c}.(1-e^{-t/RC})\:120=200(1-e^{-}\frac{5}{RC})$

$\Rightarrow 1-e^{-\frac{5}{RC}}=0.6$

$\Rightarrow e^{-5/RC}=0.40=\frac{2}{5}$

$\Rightarrow \frac{5}{Rc}=ln(5/2)$

$R=\frac{5}{ln(5/2)}.c=\frac{5}{2\times 10^{-6}\times ln5/2}=2.7\times 10^{6}\Omega$

Option 1)

$1.3 \times 10^{4}\Omega$

This option is incorrect

Option 2)

$1.7 \times 10^{5}\Omega$

This option is incorrect

Option 3)

$2.7 \times 10^{6}\Omega$

This option is correct

Option 4)

$3.3 \times 10^{7}\Omega$

This option is incorrect

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prateek

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