Question

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 $120\,V$ . Find the value of $R$ to make the bulb light up for $5\,s$ after the switch has been closed. ( ${\log _{10}}2.5 = 0.4$ )
A. $1.3 \times {10^4}\,\Omega$
B. $1.7 \times {10^5}\,\Omega$
C. $2.7 \times {10^6}\,\Omega$
D. $3.3 \times {10^7}\,\Omega$

Answer 1

When capacitor and resistance are connected together in a circuit, the resistance tries to oppose the flow of current in the capacitor. This leads to charging and discharging of the capacitor. The time constant in RC circuits indicates the time taken for charging of the capacitor. Using the relation between voltage and capacitor. Time constant in RC circuit is the product of resistance and capacitance. Substitute the value of time constant in the voltage equation to find resistance.

Complete step by step answer:
Capacitor is a device which is used to store energy. It is made up of two or more conductors which are separated by an insulator. We will first calculate the value of time constant.
Time constant $\tau$ is product of resistance $R$ and capacitance $C$ , this implies
$\tau = RC$
Now, the voltage and capacitance are related by the formula;
${V_C} = {V_S}\left( {1 - {e^{\dfrac{{ - t}}{\tau }}}} \right)$
Here, ${V_C}$ is voltage of capacitor, ${V_S}$ is the supply voltage, $e$ is rational number
$t$ is time, $\tau$ is time constant.
The given values are ${V_C} = 120\,V$ and ${V_S} = 200\,V$
$120 = 200\left( {1 - {e^{\dfrac{{ - t}}{\tau }}}} \right)$
$\Rightarrow 1 - {e^{\dfrac{{ - t}}{\tau }}} = \dfrac{3}{5}$
$\Rightarrow {e^{\dfrac{{ - t}}{\tau }}} = \dfrac{5}{2}$
Taking log, we get
$\dfrac{t}{\tau } = {\log _e}2.5$
We know that ${\log _e}a = 2.303{\log _{10}}a$
$\dfrac{t}{{RC}} = 2.303{\log _{10}}2.5$ as $\tau = RC$
Using the given value of ${\log _{10}}2.5 = 0.4$ , we get
$\dfrac{t}{{RC}} = 2.303 \times 0.4$
Substituting the value of capacitance $2\mu F$ and time as $5\,s$ , we get
$R = \dfrac{5}{{2.303 \times 0.4 \times 2 \times {{10}^{ - 6}}}}$
$\therefore R = 2.73 \times {10^6}\,\Omega$
Therefore, the value of resistance is $2.7 \times {10^6}\,\Omega$

Thus, option B is the correct option.

Note: The plates of the capacitor carry equal and opposite charges. Thus, the net amount of charge will always be zero. When the voltage is higher, the amount of energy stored in the capacitor is high. When the capacitor discharges, the potential difference between the plates disappears as the energy is transferred to the entire circuit.