Resistance in RC circuit with neon bulb – Rankers Physics

Charging and Discharging of Capacitors: Practice Problem & Solution

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. \((log_{10} 2.5 = 0.4)\)
\(2.7 \times 10^6~\Omega\)
\(3.3 \times 10^7~\Omega\)
\(1.3 \times 10^4~\Omega\)
\(1.7 \times 10^5~\Omega\)

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Charging and Discharging of Capacitors. Understanding the underlying formula is key to arriving at the correct answer below:

The charging voltage is \( V = V_0(1 - e^{-t/RC})\), so \(120 = 200(1 - e^{-t/RC})\) ⇒ \(e^{t/RC} = 2.5\). This gives \(t/RC = \ln 2.5 = 2.303 \log_{10} 2.5 \approx 0.921\). Solving with \(t = 5~text{s}\) and \(C = 2~\mu\text{F}\) gives \(R \approx 2.7 \times 10^6~\Omega\).

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