a 2 microfarad and an electrostatic voltmeter are connected to a 500 Volts supply. after disconnection the voltmeter reading falls to 250 volts in 15 minutes. find the the leakage resistance?

To find the leakage resistance, we can use the formula:

Resistance (R) = (Voltage difference) / (Current)

First, let's calculate the voltage difference:

Voltage difference = Initial voltage - Final voltage
= 500 V - 250 V
= 250 V

Next, we need to calculate the current. Since we are given the capacitance of the circuit, we can use the formula:

Current (I) = Capacitance (C) * Rate of change of voltage (dV/dt)

The rate of change of voltage can be found by dividing the voltage difference by the time it takes for the voltmeter reading to fall to 250 V:

Rate of change of voltage = Voltage difference / Time
= 250 V / (15 minutes * 60 seconds/minute)

Now, let's calculate the current:

Current = 2 microfarads * (250 V / (15 minutes * 60 seconds/minute))

Lastly, we can calculate the leakage resistance using the formula:

Resistance = Voltage difference / Current

Let's substitute the values into the formula to find the leakage resistance:

Resistance = 250 V / ((2 microfarads) * (250 V / (15 minutes * 60 seconds/minute)))

Calculating the above expression will give us the value for the leakage resistance.

To find the leakage resistance, we can use the formula:

\(R = \frac{{V_L}}{{C \cdot t}}\)

Where:
- R is the leakage resistance,
- VL is the change in voltage (initial voltage - final voltage),
- C is the capacitance, and
- t is the time taken for the voltage to change.

In this case, the initial voltage is 500 volts, the final voltage is 250 volts, and the time taken is 15 minutes. The capacitance is given as 2 microfarads (μF), but we need to convert it to farads for the formula.

1 microfarad (μF) = 1 × 10^-6 farads (F)

Therefore, 2 μF = 2 × 10^-6 F.

Now we can substitute the values into the formula to find the leakage resistance:

\(R = \frac{{500V - 250V}}{{2 \cdot 10^{-6}F \cdot 15\text{ minutes}}} \)

First, convert the time from minutes to seconds:

1 minute = 60 seconds

15 minutes = 15 × 60 seconds = 900 seconds

Now calculate:

\(R = \frac{{500V - 250V}}{{2 \cdot 10^{-6}F \cdot 900\text{ seconds}}} \)

\(R = \frac{{250V}}{{2 \cdot 10^{-6}F \cdot 900\text{ seconds}}} \)

Simplifying further:

\(R = \frac{{250V}}{{1.8 \times 10^{-3}F \cdot \text{ seconds}}} \)

\(R = \frac{{250V}}{{1.8 \times 10^{-3}C \cdot \text{ s}}} \)

\(R = \frac{{250V}}{{1.8 \times 10^{-3}A \cdot \text{ s}}} \) (Since Coulomb = Ampere-second)

\(R = \frac{{250V}}{{1.8 \times 10^{-3}A \cdot \text{ s}}} \)

\( R \approx 138,889 \, \Omega\) (rounded to the nearest ohm)

Therefore, the leakage resistance is approximately 138,889 ohms.

initial Vi = 500

initial Qi = C Vi = 2*10^-6 *500 = 10^-3 coulombs
that leaks out with current i = -dQ/dt
V(t) = i R = Q /C
so
-dQ/dt = Q/RC
dQ/Q = -(1/RC) dt
Q = constant e^-t/RC = constant e^-t/RC
at t = 0 , Q = 10^-3 so
Q = 10^-3 e^-t/RC
in 15 min = 900 seconds
Q = 10^-3 e^-900/RC
but V = Q / C
so
250*2*10^-6 = 10^-3 e*-900/RC
5*10^-4 = e^-900/(2*10^-6 R)
ln 0.0005 = -900/(2*10^-6R)
-7.6 R =- 900*10^6 /2