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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=

Rl=

^{10*14 }x^{12 x 8.85*-12 }/^{100*-12}**3.**

^{Rl=1.062*15}surley the leakage resistance cant be that high my freind 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.