Answer to Question #102976 in Electric Circuits for princess24

Question

The RC circuit below includes a 12 V battery, a 60Ω resistor and a 10 mF capacitor. (a) Show the complete calculation of the initial charge current in the circuit. (b) Show the complete calculation of the time constant for charging the capacitor. (b) Show the complete calculation of the charge current after 6.0 s. (c) Show the complete calculation of the final charge on the capacitor.

Answer 1

As per the given question,

The emf of the battery ($\epsilon$ )=12V

Resistance R=60$\Omega$

Capacitance of the capacitor c=10mF

a)

At the time t=0,

Capacitor will be uncharged, so charge on the capacitor =0

When the capacitor will get fully charged, then

$i=\dfrac{\epsilon}{R}=\dfrac{12}{60}=0.2A$

b)

Now applying the kirchoff rule

$\epsilon =iR+\dfrac{Q}{c}$

$Q$ = stands for the charge on capacitor

i= stands for the current in the circuit.

Now,

$R\dfrac{dQ}{dt}=\epsilon+\dfrac{Q}{c}$

$\dfrac{dQ}{C\epsilon -Q}=\dfrac{dt}{RC}$

$\dfrac{dQ}{Q-C\epsilon}=-\dfrac{dt}{RC}$

taking the integration of both side,

$\int_{0}^Q\dfrac{dQ}{Q-C\epsilon }=-\int_0^t\dfrac{dt}{RC}$

$[\ln{Q-C\epsilon}]_{o}^Q=-[\dfrac{t}{RC}]_0^t$

$\ln{(Q-C\epsilon)}-\ln{C\epsilon}=-\dfrac{t}{RC}$

$\ln{\dfrac{Q-C\epsilon}{C\epsilon}}=-\dfrac{t}{RC}$

Now, taking antilog,

$\dfrac{Q-C\epsilon}{C\epsilon}=e^\dfrac{-t}{RC}$

$Q-C\epsilon=C\epsilon e^\dfrac{-t}{RC}$

$Q=C\epsilon+C\epsilon e^\dfrac{-t}{RC}$

Here $RC$ is called the time constant,

$RC=60\times 10mF=600\times10^{-3}=0.6sec$

c)

Charge current after the time t=0.6

Intital charge is zero, so final charge will be

$Q=C\epsilon e^{\dfrac{-0.6}{0.6}}=\dfrac{c\epsilon}{e}=\dfrac{0.12}{2.71}=0.044c$

d)

At t= infinite,

capacitor will get fully charged,

So, charge on the capacitor Q=$cV=10\times 12\times 10^{-3}=0.12C$