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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)


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





Potential difference across capacitor


\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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