Question
Dielectric materials used in the manufacture of capacitors are characterized by conductivities that are small but not zero. Therefore, a charged capacitor slowly loses its charge by "leaking" across the dielectric. If a capacitor having capacitance $C$ leaks charge such that the potential difference decreases to half its initial value in a time $t$, what is the equivalent resistance of the dielectric?
Step 1
This can be represented by the equation: \[V = V_0 e^{-t/\tau}\] where \(V\) is the instantaneous potential, \(V_0\) is the initial potential, \(t\) is the time, and \(\tau\) is the time constant of the RC circuit. Show more…
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Dielectric materials used in the manufacture of capacitors are characterized by conductivities that are small but not zero. Therefore, a charged capacitor slowly loses its charge by "leaking" across the dielectric. If a capacitor having capacitance $C$ leaks charge such that the potential difference has decreased to half its initial $(t=0)$ value at a time $t,$ what is the equivalent resistance of the dielectric? FIGURE CANT COPY
Dielectric materials used in the manufacture of capacitors are characterized by conductivities that are small but not zero. Therefore, a charged capacitor slowly loses its charge by "leaking" across the dielectric. If a certain $3.60-\mu \mathrm{F}$ capacitor leaks charge such that the potential difference decreases to half its initial value in $4.00 \mathrm{~s}$, what is the equivalent resistance of the dielectric?
Let $C$ be the capacitance of a capacitor discharging through a resistor $R$. Suppose $t_{1}$, is the time taken for the energy stored in the capacitor to reduce to half its initial value and $t_{2}$ is the time taken for the charge to reduce to one-fourth its initial value. Then, the ratio $\underline{t_{1}}$ will be [AIEEE 2010] $t_{2}$ (a) 1 (b) $\frac{1}{2}$ (c) $\frac{1}{4}$ (d) 2
Electrostatics
Round 2
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