In the loss of charge method unknown resistance is connected in parallel with the capacitor and electrostatic voltmeter. The capacitor is initially charged to some suitable voltage by means of a battery of voltage V and then allowed to discharge through the resistance. The terminal voltage is observed during discharge and it is given by,

OR

Or insulation resistance is given by,

The variation of voltage v with time is shown in figure,

From above equation, it follows that if V, v, C, and t are known the value of R can be computed.

If the resistance R is very large the time for an appreciable fall in voltage is very large and thus this process may become time-consuming. Also the voltage-time curve will thus be very flat and unless great care is taken in measuring voltages at the beginning and at the end of time t, a serious error may be made in the ratio V/v causing the considerable corresponding error in the measured value of R. more accurate results may be obtained by change in the voltage V-v directly and calling this change as e, the expression for R becomes:

This change in voltage may be measured by a galvanometer.

However, from the experimental point of view, it may be advisable to determine the time t from the discharge curve of the capacitor by plotting the curve of log v against time t. this curve is linear as shown in the second figure and thus the determination of time t from this curve for the voltage to fall from V to v yields more accurate results.

Loss of charge method is applicable to some high resistances, but it requires a capacitor of very high leakage resistance as high as resistance being measured. The method is very attractive if the resistance being measured is the leakage resistance of a capacitor as in this case auxiliary R and C units are not required.

Actually, in this method, we do not measure the true value of resistance since we assume here that the value of resistance of electrostatic voltmeter and the leakage resistance of the capacitor have infinite value. But in practice corrections must be applied to take into consideration the above two resistances. Let R1 be the leakage resistance of the capacitor. Also R’ be the equivalent resistance of the parallel resistances R and R1.

Then discharge equation of capacitor gives,

R’=0.4343 t / (C log V/v)

The test is then repeated with the unknown resistance R disconnected and the capacitor discharging through R1. The value of R1 obtained from this second test and substituted into the expression,

R’=(R R1) / (R+R1)

In order to get the value of R.

The leakage resistance of the voltmeter, unless very high should also be taken into consideration.

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give a example for loos and charge

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