How Do You Calculate the Leakage Resistance of a Capacitor?

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SUMMARY

The leakage resistance of a capacitor can be calculated using the formula Rl = ρ * Er * Eo / C. For a capacitor with a capacitance of 100pF, relative permittivity of 12, and resistivity of 10^14 ohm-meters, the calculated leakage resistance is approximately 1076.87 ohms. This value, while significant, reflects the ideal behavior of a capacitor, which theoretically has infinite leakage resistance. The calculations confirm that the leakage resistance can be substantial, but practical measurements may not capture such high values.

PREREQUISITES
  • Understanding of capacitor specifications, including capacitance and relative permittivity.
  • Familiarity with electrical resistivity and its units (ohm-meters).
  • Knowledge of fundamental electrical formulas, particularly those involving capacitance and resistance.
  • Basic skills in scientific notation and unit conversions.
NEXT STEPS
  • Research the impact of leakage resistance on capacitor performance in electronic circuits.
  • Explore the characteristics of ideal versus real capacitors and their implications in circuit design.
  • Learn about measuring techniques for high resistance values using specialized equipment.
  • Investigate the effects of temperature and material properties on the leakage resistance of capacitors.
USEFUL FOR

Electrical engineers, electronics students, and anyone involved in capacitor design or analysis will benefit from this discussion, particularly those focusing on leakage resistance and its implications in circuit performance.

tommoturbo
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1. Calculate the leakage resistance across a capacitor of 100pF that has a relative permittivity of 12 and a resistivitty of rho=10*14



2. Rl=rho x Er x Eo /C
Rl=10*14 x 12 x 8.85*-12 / 100*-12




3. Rl=1.062*15


surley the leakage resistance can't be that high my friend said it should be this formula

You need to use the formula; R = rho E/C (it's the lesson books somewhere!), where E = ErEo.

So, E = ErEo which is; 12*8.85*10^-12. This equals 1.062*10^-10

Therefore R = 1014 * 1.062*10^-10/100*10^-12.

This gives R as 1076.87 ohms.
 
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I calculated

R_{leak}=1.062\times10^{14} ohms

Sure it's a big value - not something you measure with a multimeter! - but an ideal capacitor would have infinite leakage resistance.
 

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