The above circuit is built, and is powered by a 10 V battery. If the capacitance

Q: The above circuit is built, and is powered by a 10 V battery. If the capacitance

The current through the resistor in RC circuit is given as I = (V/R) e^(-t/RC) I = (10/41.25)e^(-5/(41.25)(30x10^-6)) Energy dissipated through the resistor in first five second is E = I^2 R =R [(10/41.25)e^(-5/(41.25)(30x10^-6))]^2 This is too small as the energy almost given to the capacitor to charge.

ID: 1260187 • Letter: T

Question

The above circuit is built, and is powered by a 10 V battery. If the capacitance of the capacitor is 20 microfarads, the resistance of each resistor is 30 ohms, and the toal resistance is 41.25 ohms, derive how much energy is delivered to the total resistance within the first five seconds after connecting the power supply.

Explanation / Answer

The current through the resistor in RC circuit is given as

I = (V/R) e^(-t/RC)

I = (10/41.25)e^(-5/(41.25)(30x10^-6))

Energy dissipated through the resistor in first five second is

E = I^2 R =R [(10/41.25)e^(-5/(41.25)(30x10^-6))]^2

This is too small as the energy almost given to the capacitor to charge.

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The above circuit is built, and is powered by a 10 V battery. If the capacitance

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