Question

A neon gas tube is connected in parallel to a 1 $mF$ capacitor, which is charged by a source of emf 15 V through a 1 $k\Omega$ resistor as shown. The switch is closed at $t=0s$. When the potential difference across the tube reaches the firing value $V_F=10V$, neon gas starts conducting with almost zero resistance and emits a flash of light. The capacitor then discharges rapidly and when the potential drops to extinction V${}_E=5\vee$, the gas becomes an insulator, and the capacitor again starts charging. The period of flashing $\tau$ is about

• A. $300ms$
• B. $500ms$
• C. $100ms$
• D. $700ms$

Answer 1

The correct option is D $700ms$

$V_0=15V,V_F=10V,V_E=5V,R=1000\Omega ,C=1mF$

$V_E=V_0(1-e^{-t_0/RC}).....\left(1\right)$

$V_F=V_0(1-e^{-(t_0+\tau )/RC}).....\left(2\right)$

from $\left(1\right),e^{-t_0/RC}=\frac{V_0-V_E}{V_0}$

from $\left(2\right),e^{-(t_0+\tau )/RC}=\frac{V_0-V_F}{V_0}$

or,$\frac{V_0-V_E}{V_0}{e}^{-\tau /RC}=\frac{V_0-V_F}{V_0}$

$\frac{(V_0-V_E)}{(V_0-V_F)}=e^{\tau /RC}$

$\therefore \tau =RCln\frac{(V_0-V_E)}{(V_0-V_F)}={10}^3\times {10}^{-3}ln\left(\frac{10}{5}\right)=1\times ln2$

$=0.69s\simeq 0.7s=700ms$