Since you analyzed the charging of a capacitor for a DC charging voltage, how is it possible that you August 13, 2021 by Minh Khuê Since you analyzed the charging of a capacitor for a DC charging voltage, how is it possible that you

Answer: I = E/R e^{-t/RC} Explanation: In a capacitor charging circuit you must have a DC power source, the capacitor, a resistor, and a switch. When closing the circuit, E -q / c-IR = 0 we replace the current by its expression and divide by the resistance I = dq / dt dq / dt = E / R -q / RC dq / dt = (CE -q) / RC we solve the equation dq / (Ce-q) = -dt / RC we integrate and evaluate for the charge between 0 and q and for the time 0 and t ln (q-CE / -CE) = -1 /RC (t -0) eliminate the logarithm q – CE = CE [tex]e^{-t/RC}[/tex] q = CE (1 + 1/RC e^{-t/RC} ) In general the teams measure the current therefore we take the derivative to find the current i = CE (e^{-t/RC} / RC) I = E/R e^{-t/RC} This expression is the one that describes the charge of a condensate in a DC circuit Reply

Answer:I = E/R e^{-t/RC}

Explanation:In a capacitor charging circuit you must have a DC power source, the capacitor, a resistor, and a switch. When closing the circuit,

E -q / c-IR = 0

we replace the current by its expression and divide by the resistance

I = dq / dt

dq / dt = E / R -q / RC

dq / dt = (CE -q) / RC

we solve the equation

dq / (Ce-q) = -dt / RC

we integrate and evaluate for the charge between 0 and q and for the time 0 and t

ln (q-CE / -CE) = -1 /RC (t -0)

eliminate the logarithm

q – CE = CE [tex]e^{-t/RC}[/tex]

q = CE (1 + 1/RC e^{-t/RC} )

In general the teams measure the current therefore we take the derivative to find the current

i = CE (e^{-t/RC} / RC)

I = E/R e^{-t/RC}

This expression is the one that describes the charge of a condensate in a DC circuit