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Oxana
1 month ago
6

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?
Physics
2 answers:
Maru [3.3K]1 month ago
6 0

Response:

I = 0.667 A

Clarification:

At the moment immediately after the switch closes, the capacitor essentially acts like a short circuit; therefore, we can ignore it in our calculations. Resistance is measured in watts or equivalent power units using current and voltage, and the ammeter also behaves like a short circuit.

Taking all this into account, the circuit can be simplified to a battery connected in series with a resistor, allowing us to use the power formula to solve the issue as follows:

P = I*V

4W = I*6V

I = 0.667 A

Softa [3K]1 month ago
4 0
<span>You are presented with a circuit that includes a 6.0-v battery, a 4.0-ohm resistor, a 0.60 microfarad capacitor, an ammeter, and a switch all connected in series. Your task is to determine the current reading once the switch is closed. Ohm's law should be used, which states V = IR where V signifies voltage, I indicates current, and R represents resistance.</span>

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

Upon closing the switch, the cathode side plate starts accumulating electrons if it was previously empty. As this process continues, the current diminishes. Eventually, when the capacitor reaches its maximum electron retention, the current will cease. An increased capacitance means a greater capacity for electron storage.
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Answer:

7.166 hours = 430 minutes.

Explanation:

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X ----> 860 miles; hence X = (860 miles * 1 hour)/120 miles = 43/6 hours = 7.16666 hours. To convert this into minutes, recall that 1 hour equals 60 minutes; therefore, 43/6 hours * 60 minutes/hour = 430 minutes.

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Details:

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2 months ago
Read 2 more answers
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