# RC circuit behavior at t=0

A $2.36 \mu F$ that is initially uncharged is connected in series with a $58.6 \Omega$ resistor and an emf source of $120 V$ and an internal resistance of  $3 \Omega$.

Just after the connection is made, what are:

I) The rate at which electrical energy is being dissipated in the resistor.

II) The rate at which the electrical energy stored in the capacitor is increasing.

III) The electrical power output of the source.

Initially the current in circuit is maximum (the capacitor is not charged and acting like a short circuit)

$I_max = U/(R_1+R_2) =120/(58.6 +3) =1.948 A$

$P(R) = I^2*R =1.948^2*58.6 =222.38 W$

Voltage on capacitor is

$V= V_{max}(1-exp(-t/\tau))$ so that for

$V max =U =120 V$ we have  $\tau = (R_1+R_2)*C =(58.6+3)*2.36*10^{-6} =0.14538 ms$

Energy stored on capacitor

$E(t) = C*U^2/2 = (C*V_{max}^2)/2 * (1-exp(-t/\tau))^2$

for $t=0$ we have $E =0$ and thus

$P =d E/d t = [(C*U^2)/2] * 2*exp(-t/\tau) *(1-exp(-t/\tau))/\tau >0$

Which means the energy on capacitor is increasing with time

The rate of energy to time

$d P/d t = d^2E/d t^2 = (CU^2)/\tau * [exp(-2t/\tau) -exp(-t/\tau)*(1-exp(-t/\tau)] =$

$=(CU^2)/\tau *exp(-t/\tau)*[exp(-t/\tau) -1 + exp(-t/\tau)] = – (CU^2)/\tau * exp(-t/\tau) < 0$

Which means the RATE OF ENERGY with time is decreasing

$I = I_{max}*exp(-t/\tau)$

Electrical power output of source is

$P(source) =I^2*R(source) = I_{max}^2 *3 *exp(-t/\tau) =$

$=1.948^2*3 *exp(-t/\tau) =11.38 *exp(-t/\tau) W$

and thus

at $t= 0$

$P(source) = P_{max}= 11.38 W$