This problem involves an RC circuit where a resistor (R) and capacitor (2μF) are connected in series to a 200V DC supply. A neon bulb, connected across the capacitor, lights up when the capacitor voltage reaches 120V. We need to find the value of R such that this happens 5 seconds after closing the switch. The hint given is log₁₀(2.5) = 0.4.

Step 1: Understand the Charging Process

When the switch is closed, the capacitor starts charging through the resistor. The voltage across the capacitor (V_C) at any time 't' is given by the charging equation:

$V_C=E(1-e^{-\frac{t}{RC}})$

Where:

• E = Supply voltage = 200 V
• R = Resistance (unknown, to be found)
• C = Capacitance = 2 μF = 2 × 10^-6 F
• t = Time = 5 s
• V_C = Voltage at which bulb lights = 120 V

Step 2: Plug in the Known Values

Substitute the known values into the equation:

$120=200(1-e^{-\frac{5}{R\times 2\times {10}^{-6}}})$

Step 3: Solve for the Exponential Term

First, simplify the equation:

$\frac{120}{200}=1-e^{-\frac{5}{2R\times {10}^{-6}}}$

$0.6=1-e^{-\frac{5}{2R\times {10}^{-6}}}$

Now, rearrange to isolate the exponential term:

$e^{-\frac{5}{2R\times {10}^{-6}}}=1-0.6=0.4$

Step 4: Take Natural Logarithm (ln) on Both Sides

To solve for the exponent, we take the natural logarithm:

$-\frac{5}{2R\times {10}^{-6}}=\ln \left(0.4\right)$

We know that ln(0.4) = -ln(2.5) because 1/0.4 = 2.5. Therefore:

$-\frac{5}{2R\times {10}^{-6}}=-\ln \left(2.5\right)$

The negative signs cancel out:

$\frac{5}{2R\times {10}^{-6}}=\ln \left(2.5\right)$

Step 5: Use the Given Logarithmic Value

The problem gives log₁₀(2.5) = 0.4. We need to convert this to a natural logarithm (ln). The conversion is: ln(x) = 2.3026 × log₁₀(x).

$\ln \left(2.5\right)=2.3026\times 0.4\approx 0.92104$

So our equation becomes:

$\frac{5}{2R\times {10}^{-6}}=0.92104$

Step 6: Solve for R

Rearrange the equation to solve for R:

$R=\frac{5}{2\times 0.92104\times {10}^{-6}}$

First, calculate the denominator:

$2\times 0.92104\times {10}^{-6}=1.84208\times {10}^{-6}$

Now, calculate R:

$R=\frac{5}{1.84208\times {10}^{-6}}\approx 2.714\times {10}^6 \Omega$

This is approximately 2.7 × 10⁶ Ω.

Final Answer: The value of R is $2.7\times {10}^6 \Omega$

Related Topics & Formulae

RC Circuits: An RC circuit is composed of resistors and capacitors. The time constant (τ = R×C) determines how quickly the capacitor charges or discharges.

Capacitor Charging Equation: The voltage across a charging capacitor at time 't' is given by $V_C=E(1-e^{-\frac{t}{RC}})$, where E is the source voltage.

Logarithmic Relationships: Solving exponential equations often requires using natural logarithms (ln) or common logarithms (log₁₀), with the conversion ln(x) ≈ 2.3026 × log₁₀(x).