Question

An initially uncharged 3.67 μF capacitor and a 8.01 k Ω resistor are connected in series to a 1.50 V battery that has negligible internal resistance. What is the initial current in the circuit, expressed in milliamperes?

Answer 1

Answer:

Explanation:

Given an RC series circuit

Initially uncharged capacitor

C=3.67 μF

Resistor R=8.01 k Ω=8010 ohms

Battery EMF(V)=1.5V with negligible internal resistance.

The initial current in the circuit?

At the beginning the capacitor is uncharged and it has a 0V, so all the voltage appears at the resistor,

Now using ohms law

V=iR.

i=V/R

i=1.5/8010

i=0.000187A

1mA=10^-3A

Therefore, 1A = 1000mA

i=0.187 milliamps

The initial current in the circuit is 0.187 mA

Answer 2

Answer:

The initial current is 0.0187 mA.

Explanation:

Given that

capacitance is given as 3.67 x 10⁻⁶ F

resistance is given as 8010 Ω

voltage across the circuit is 1.5 V

Since the capacitor is initially uncharged, the capacitive reactance is zero.

From ohms law;

Voltage across the circuit is directly proportional to the opposition to the flow of current.

In these circumstances, as the battery only "sees"a resistor, the initial current can be found applying Ohm's law to the resistor, as follows:

$V = I_{0}*R \\\\ I_{0} = \frac{V}{R} = \frac{1.50V}{8.011e3\Omega}\\ = 0.0187 mA$

The initial current (that will be diminishing as the capacitor charges), is 0.0187 mA.