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1. A series connection of a 2 micro-F capacitor and a 5 mega-ohm resistor is connected...
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At t=0, the capacitor is uncharged and the current can be found using ohm’s law.
As the capacitor begins to accumulate charge, the current flow begins to decrease and ceases completely when the capacitor is fully charged (because the resistor and capacitor are in series).
The voltage and current at time t can be obtained from the charging equation of a resistor-capacitor circuit which is given by:
Charging, `V = V_0*(1-e^(-t/t_c)) `
`i(t) = V_0/R*e^(-t/t_c)` where, `t_c` is the time constant of the circuit.
Here, `t/t_c` = 1 (since one time constant has elapsed),
C=2 *10^-6 F
R = 5*10^6 ohms
V = 100 Volts,
Therefore, `i = 100*(e^-1)/(5*10^6)`
= `7.3576 * 10^-6 ` Amperes.
Therefore, the current in the circuit after one time constant has elapsed is 7.3576 micro amperes.
Posted by llltkl on June 21, 2013 at 6:38 PM (Answer #1)
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