Question

An initially uncharged 4.07 μF capacitor and a 7.71 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:

I₀ = 0.2 mA

Explanation:

• Just after being connected, as the voltage between plates of a capacitor can't change instantanously, the initial voltage through the capacitor must be zero, so it presents like a short to the battery.
• So, 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:

        $V = I_{0}*R \\\\ I_{0} = \frac{V}{R} = \frac{1.50V}{7.71e3\Omega} = 0.2 mA$

• The initial current (that will be diminishing as the capacitor charges), is 0.2 mA.

Answer 2

Answer:

0.19mA

Explanation:

Given;

Capacitor of capacitance = 4.07 μF

Resistor of resistance = 7.71 kΩ = 7.71 x 1000Ω  = 7710Ω

Voltage = 1.50V

Since the capacitor is initially uncharged, it behaves like a short circuit. Therefore, the only element drawing current at that instant is the resistor.  This means that the initial current in the circuit is the one due to (flowing through) the resistor.

And since there is negligible internal resistance, the emf of the battery is equal to the voltage supplied by the battery and is used to supply current to the resistor. Therefore, according to Ohm's law;

V = I x R           ---------------(i)

Where;

V = voltage supplied or the emf

I = current through the resistor

R = resistance of the resistor

Substitute the values of V and R into equation (i) as follows;

1.50 = I x 7710

I = $\frac{1.5}{7710}$

I = 0.00019A

Multiply the result by 1000 to convert it to milliamperes as follows;

0.00019 x 1000 mA = 0.19mA

Therefore, the initial current in the circuit, expressed in milliamperes is 0.19