Problem 134 - Master Resource Book in JEE Main Physics

Question

A resistor \(R\) and \(2 \mu \mathrm{F}\) capacitor in series is connected through a switch to \(200 \mathrm{~V}\) direct supply. Across the capacitor is a neon bulb that lights up at \(120 \mathrm{~V}\). Calculate the value of \(R\) to make the bulb light up \(5 \mathrm{~s}\) after the switch has been closed \(\left(\log _{10} 2.5=0.4\right)\) [AIEEE 2011] (a) \(1.7 \times 10^{5} \Omega\) (b) \(2.7 \times 10^{6} \Omega\) (c) \(3.3 \times 10^{7} \Omega\) (d) \(1.3 \times 10^{4} \Omega\)

Step-by-Step Solution

Answer

The resistor value to light up the bulb in 5s is approximately \(4.888 \times 10^6 \Omega\).

Step 1: Understand the circuit and the initial condition

This circuit contains a resistor and capacitor in series connected to a 200 V DC supply with a switch. The capacitor charges through the resistor until the voltage across it reaches 120 V, which is the threshold for the neon bulb to light up.

Step 2: Use the charging formula for capacitors

The capacitor charges according to the formula \( V(t) = V_0(1 - e^{-t/RC}) \), where \( V_0 = 200 \) V, \( V(t) = 120 \) V, \( t = 5 \) seconds, and \( C = 2 \, \mu F \). We need to solve for the resistance \( R \).

Step 3: Substitute known values into the formula

Replace into the formula: \( 120 = 200(1 - e^{-5/(R \cdot 2 \times 10^{-6})}) \). Solve for the exponent: \( \frac{120}{200} = 1 - e^{-5/(R \cdot 2 \times 10^{-6})} \).

Step 4: Rearrange the equation

Rearranging gives \( 0.4 = 1 - e^{-5/(R \cdot 2 \times 10^{-6})} \) or \( e^{-5/(R \cdot 2 \times 10^{-6})} = 0.6 \).

Step 5: Solve for the exponent

Take the natural logarithm on both sides to isolate the exponent: \( -\frac{5}{R \cdot 2 \times 10^{-6}} = \ln(0.6) \). Calculating the natural logarithm, \( \ln(0.6) \approx -0.5108 \).

Step 6: Calculate R

Substitute \( \ln(0.6) \) into the equation to solve for \( R \): \( -\frac{5}{R \cdot 2 \times 10^{-6}} = -0.5108 \). This gives \( R \cdot 2 \times 10^{-6} = \frac{5}{0.5108} \). Calculate this value. \( R = \frac{5}{0.5108 \times 2 \times 10^{-6}} \approx 4.888 \times 10^{6} \Omega \).

Key Concepts

Capacitor Charging

Capacitor charging in an RC circuit is an exciting process that involves the gradual growth of voltage across the capacitor over time. When you close the switch, the electric field begins to store charge on the capacitor plates. The rate at which this happens is governed by the time constant of the circuit, which is the product of the resistance \( R \) and capacitance \( C \).
\[ \tau = R \cdot C \]
When the capacitor is charging, the voltage across it follows a specific time-dependent formula:
\[ V(t) = V_0(1 - e^{-t/RC}) \]
Where:

• \( V_0 \) is the maximum voltage supplied by the power source, in this case, 200 V.
• \( t \) is the time elapsed since the switch was closed.
• \( RC \) is the time constant describing the speed of charging.

In our problem, the goal is to know when the voltage reaches 120 V—enough to light up the neon bulb. As you can see, understanding the charging equation allows us to model how quickly or slowly the capacitor will charge, which is crucial for timing circuits or applications involving precise voltage levels.

Neon Bulb Ignition

A neon bulb ignites when the voltage across it exceeds its break-even or firing voltage. For this example, the neon bulb will light up when the voltage reaches 120 V. This process is very interesting in electronics since such a bulb acts like a voltage sensor!

Initially, the neon bulb remains off until the necessary voltage level appears across the capacitor. When the required 120 V is reached, the gas inside the bulb becomes ionized, allowing current to pass through it and causing the bulb to glow.
If we hadn't used a neon bulb in our circuit, the capacitor could charge up to the full 200 V without an indication. Therefore, neon bulbs are widely used in circuits for indicating when a particular voltage level has been achieved.

In the context of the exercise, the bulb acts as our indicator, telling us exactly when the capacitor has charged to 120 V, right at the moment the bulb starts glowing.

Resistor Calculation

Determining the right resistor is key when you want the neon bulb to ignite exactly at 5 seconds after closing the switch. The resistor controls the rate at which the capacitor charges by limiting the current flow. To find this resistor value, we use the capacitor's charging equation and solve for \( R \).

The voltage equation given as:
\[ 120 = 200(1 - e^{-5/(R \cdot 2 \times 10^{-6})}) \]
Opening up this equation, you realize it's all about rearranging and using logarithms:
First, we simplify:
\[ \frac{120}{200} = 1 - e^{-5/(R \cdot 2 \times 10^{-6})} \]
This becomes:
\[ 0.4 = 1 - e^{-5/(R \cdot 2 \times 10^{-6})} \]
Rearranging gives:
\[ e^{-5/(R \cdot 2 \times 10^{-6})} = 0.6 \]
Taking the natural logarithm of both sides, we find:
\[ -\frac{5}{R \cdot 2 \times 10^{-6}} = \ln(0.6) \approx -0.5108 \]
Solving for \( R \):
\[ R \cdot 2 \times 10^{-6} = \frac{5}{0.5108} \]
This gives us the resistance needed to meet our specific timing requirement. Calculating it out shows a resistance value crucial for getting the neon bulb to light at precisely 5 seconds—a perfect match for engineering specific timing in electronic circuits.