Animated Solution for Physics - Current Electricity: A resistor $R$ and $2\mu \text{F}$ capacitor in series is connected through a switch to $200\text{ V}$ direct supply. Across the capacitor is a neon bulb that lights up at $120\text{ V}$. Calculate the value of $R$ to make the bulb light up $5\text{ s}$ after the switch has been closed (take ${\log }_{10}2.5=0.4$)

A Circuit in Waiting

Imagine a circuit where a resistor $R$ and a capacitor $C$ are connected in series with a direct current battery $E$. In parallel with the capacitor, we have a neon bulb. This bulb is a fascinating component; it acts like an open switch until the voltage across it reaches a specific threshold. In our case, this threshold is $120\text{ V}$.

Because the neon bulb is filled with gas, its initial resistance is practically infinite. This means that before it lights up, no current flows through the bulb's branch. The circuit behaves exactly like a simple, standard RC charging circuit.

Building Up the Voltage

As soon as the switch is closed, the battery starts pushing charge onto the capacitor plates through the resistor. The voltage across the capacitor doesn't jump instantly; it builds up over time according to the classic charging equation:

We are given that the battery voltage $E$ is $200\text{ V}$. The bulb lights up when the capacitor voltage $V_c$ reaches $120\text{ V}$ exactly at time $t=5\text{ s}$. Let's substitute these known values into our equation to see what we get:

Solving for Resistance

Now, the goal is to isolate our unknown variable, the resistance $R$. First, let's simplify the equation by dividing both sides by $200$:

This simplifies neatly to $\frac{3}{5}$.

Rearranging the terms to isolate the exponential part, we move the exponential to the left and $\frac{3}{5}$ to the right:

To bring the variables down from the exponent, we need to take the natural logarithm ($\ln$) of both sides.

The negative signs on both sides cancel out beautifully, leaving us with:

Rearranging this to solve for $R$, we get:

Calculating the Value

We are given the value of the base-10 logarithm, ${\log }_{10}(2.5)=0.4$. To use this, we must convert our natural logarithm to a base-10 logarithm using the conversion factor $2.303$:

Now, let's substitute all our values, including the capacitance $C=2\mu \text{F}=2\times 10^{-6}\text{ F}$:

Let's crunch the numbers in the denominator: $2\times 0.4=0.8$, and $0.8\times 2.303\approx 1.8424$.

Dividing $5$ by $1.8424$ gives approximately $2.71$. The $10^{-6}$ in the denominator moves to the numerator as $10^6$.

And there we have it! The required resistance is $2.7\times 10^6\Omega$, which matches option (b). This setup is a classic example of a relaxation oscillator, where the RC time constant dictates the rhythm of the flashing bulb.