Question

A circuit contains a 6.0-v battery, a 4.0-w resistor, a 0.60-µf capacitor, an ammeter, and a switch all in series. what will be the current reading immediately after the switch is closed?

Answer 1

Answer:

I = 0.667 A

Explanation:

A capacitor behaves like a short circuit at t = 0 immediately after the switch is closed, therefore we can discard it from the equation, the resistance is given in watts or power units equivalent to current by voltage and the ammeter behave like a short circuit by nature.

Considering all of the above, the circuit looks like a battery in series with a resistor, therefore, we can use the power equation to solve the problem as follow:

P = I*V

4W = I*6V

I = 0.667 A  

Answer 2

You are given a circuit that contains a 6.0-v battery, a 4.0 ohm resistor, a 0.60 micro farad capacitor, an ammeter, and a switch all in series. You are asked to find the current reading after the switch is closed. Apply ohms law where V = IR where V is the voltage, I is the current and R is the resistor.

V = IR
I = V/R
I = 6 volts / 4 ohms
I = 1.5A

When the switch is closed, the cathode side plate begins to fill up with electrons when it was originally empty before the switch was closed. When it fills up the cathode side of the circuit, the current decreases. And when the capacitor cannot hold more electrons, the current will stop. The higher the capacitance, the higher is the capacity to store electrons.