# A Few RC Circuits/Electricity Problems

• cse63146
In summary, the conversation discusses various equations and calculations related to electric current, electric field strength, and charge. These include determining the electric field strength needed to create a 5.0 A current in a 2.0-mm-diameter iron wire, designing a 1.0 A fuse that "blows" if the current exceeds 1.0 A, calculating the amount of charge that leaves a car battery when it is completely discharged, and finding the expression for current in a wire at a given time. The conversation also mentions the equations E = \frac{I}{(\sigma)(\pi)r^2}, J = \frac{I}{(\pi)(r^2)}, and Q = \frac{I}{t
cse63146

## Homework Statement

1. What electric field strength is needed to create a 5.0 A current in a 2.0-mm-diameter iron wire?

2. You need to design a 1.0 A fuse that "blows" if the current exceeds 1.0 A. The fuse material in your stockroom melts at a current density of 500 A/cm^2. What diameter wire of this material will do the job?

3.A car battery is rated at 90 Ahr, meaning that it can supply a 90 A current for 1 hr before being completely discharged. If you leave your headlights on until the battery is completely dead, how much charge leaves the battery? Express your answer in Coulombs to three significant figures

4. The total amount of charge in coulombs that has entered a wire at time t is given by the expression Q = 4t - t^2 , where t is in seconds and t>=0.

Which is an expression for the current in the wire at time t.

http://img100.imageshack.us/img100/5448/currentequationsgd2.jpg

## Homework Equations

E = $$\frac{I}{(\sigma)(\pi)r^2}$$

J = $$\frac{I}{(\pi)(r^2)}$$

Q = $$\frac{I}{t}$$

## The Attempt at a Solution

1. I know $$\sigma$$ is 10^7 so

E = $$\frac{5}{(10^7)(\pi)(0.001^2)}$$ = 0.159 N/c

2. 500x10^-2=5 A/m^2

r = $$\sqrt{\frac{I}{(\pi)J}}$$ = $$\sqrt{\frac{1}{(\pi)5}}$$ = 0.252 m and to get diameter I multiply it by two to get 0.504. The question asks me to express it in mm so I got 504mm.

3. 1 hour = 3600 seconds

Q = $$\frac{I}{t}$$ = Q = $$\frac{90}{3600}$$ = 0.025 C

4. Current is the derivative of charge so it would be the third equation.

Did I make a math (or any other kind) error?

Last edited by a moderator:
#1 and #4 are correct, but I cannot get the correct answers for #2 and #3. Does anyone know why?

If you have 500 amps for every square CENTImeter, should you have a lot less per square meter, or, say, a LOT more?

#3, assuming you did it right, has only 2 significant figures

it's always the easy ones ^_^

so 1 Square Centimeter = 0.0001 Square Meters therefore 500 A per Square Centimeters = 0.05 A per Square Meters

r = $$\sqrt{\frac{I}{(\pi)J}}$$ = $$\sqrt{\frac{1}{(\pi)0.05}}$$ = 5.046 m = 5046 mm

and to change 0.025 from 2 significant digits to 3, I'll need to use scientific notation so 0.025 = 2.50 x 10^-2

Ah, now I get it thanks. Let's give this another try:

r = $$\sqrt{\frac{I}{(\pi)J}}$$ = $$\sqrt{\frac{1}{(\pi)(5000000)}}$$ = 0.000252 m and to get the diameter I multiply it by two to get 5.05 x 10^-4 m = 0.505 mm.

And I also figured out why #3 is wrong. It's not Q = I/t, it's Q = I*t so:

Q = I*t = (90)(3600) = 324000 C

Just tried them, and it said I was correct. Thank you blochwave.

Last edited:

## What is an RC circuit?

An RC circuit is a circuit that contains a resistor (R) and a capacitor (C) connected in series or in parallel. These circuits are commonly used in electronic devices to control the flow of electricity.

## What is the purpose of a resistor in an RC circuit?

The resistor in an RC circuit serves to control the flow of current and to limit the amount of voltage that can pass through the circuit. It also helps to prevent damage to the capacitor by limiting the amount of current that can flow through it.

## How does a capacitor work in an RC circuit?

A capacitor stores electrical charge and releases it when needed. In an RC circuit, the capacitor charges up when the circuit is switched on, and then discharges when the circuit is switched off. This allows for the regulation of current in the circuit.

## What is the time constant of an RC circuit?

The time constant is a measure of how quickly a capacitor charges or discharges in an RC circuit. It is calculated by multiplying the resistance (R) and capacitance (C) values together. This time constant is used to determine the behavior and response of the circuit.

## What are some common applications of RC circuits?

RC circuits are commonly used in electronic devices such as radios, televisions, and computers. They are also used in timing circuits, filters, and oscillators. Additionally, RC circuits are used in the charging and discharging of batteries and in voltage regulators.

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