Question

A resistor $R$ and $2\mu F$ capacitor are connected in series through a $200V$ direct supply. Across the capacitor there
is a neon bulb that lights up at $120V$. Find the value of $R$ to make the neon bulb light up exactly $5s$ after the switch has been closed. (Take $lo{g}_{10}\left(2.5\right)=0.4$)

• A. $1.7\times {10}^5\Omega$
• B. $2.7\times {10}^6\Omega$
• C. $3.3\times {10}^7\Omega$
• D. $1.3\times {10}^4\Omega$

Answer 1

The correct option is B $2.7\times {10}^6\Omega$

Given
$E=200V$
Potential across capacitor for bulb to glow
$V_c=120V$
As in a charging circuit,
$V_c=E(1-e^{-t/RC})$

$\Rightarrow 1-e^{-t/RC}=\frac{V_c}{E}=\frac{120}{200}=\frac{3}{5}{e}^{-t/RC}=0.4e^{t/RC}=2.5$

taking $log$ at both side,
$log\left(e^{-t/RC}\right)=lo{g}_{e}2.5\frac{t}{RC}=2.3026\left(lo{g}_{10}2.5\right)=0.92$

So, the value of $R$ to light up the bulb after $5s$ is,
$R=\frac{t}{0.92C}=\frac{5}{0.92\times 2\times {10}^{-6}}=2.7\times {10}^6\Omega$