# Parallel Circuits - the Plague!

**Now that I have you thoroughly confused, let me make things as clear as mud for you. Let's begin with a very simple circuit: 2 resistors and 1 battery. You are given the following information about the circuit:**

R

R

A

Find:

Total Voltage

Total Circuit Resistance

Total Current

And Finally, the Current through A

Now this isn't as tuff as it first looks. Let's break the problem down. We know according to Ohm's law, that if we know the resistance and current, we can find the voltage.

E

E = .2 x 50 = 10

E = 10 Volts.

Now that we know that the voltage for the entire circuit is 10 volts, let's find the total Resistance.

First, we find the reciprocals of the individual resistances:

R

R

Now we add the two reciprocals together:

.02 + .005 = .025

Finally we take the reciprocal of the sum:

1 / .025 = 40 Î©

So if the Total Voltage of the circuit is 10 Volts, and the Total Resistance = 40 Î© then by using Ohms Law again we can find the total current.

I

I = 10/40 = ¼ Ampere.

Almost finished now. So far we know:

R

R

A

V

R

and

I

Now we have at least 2 methods by which we can find the current through A

We know that the Total current is the sum of all the individual leg currents, so if we subtract the current of A

I

The other method would be by using Ohms Law. We know the resistance of R

10 Volts / 200 Î© = .05 Amperes.

Either way, our final result is A

R

_{ 1 }= 50 Î©R

_{ 2 }= 200 Î©A

_{ 1 }reads .2 Amps in current ( I=.2 )Find:

Total Voltage

Total Circuit Resistance

Total Current

And Finally, the Current through A

_{ 2 }Now this isn't as tuff as it first looks. Let's break the problem down. We know according to Ohm's law, that if we know the resistance and current, we can find the voltage.

E

_{ R1 }= I_{ R1 }x R_{ 1 }.E = .2 x 50 = 10

E = 10 Volts.

Now that we know that the voltage for the entire circuit is 10 volts, let's find the total Resistance.

First, we find the reciprocals of the individual resistances:

R

_{ 1 }= 50 ohms. 1/50 = .02R

_{ 2 }= 200 ohms. 1/200 = .005Now we add the two reciprocals together:

.02 + .005 = .025

Finally we take the reciprocal of the sum:

1 / .025 = 40 Î©

So if the Total Voltage of the circuit is 10 Volts, and the Total Resistance = 40 Î© then by using Ohms Law again we can find the total current.

I

_{ Total }= E_{ Total }/ R_{ Total }I = 10/40 = ¼ Ampere.

Almost finished now. So far we know:

R

_{ 1 }= 50 Î©R

_{ 2 }= 200 Î©A

_{ 1 }reads .2 Amps in current ( I=.2 )V

_{ Total }= 10R

_{ Total }= 40and

I

_{ Total }= ¼Now we have at least 2 methods by which we can find the current through A

_{ 2 }.We know that the Total current is the sum of all the individual leg currents, so if we subtract the current of A

_{ 1 }from the Total current we get this:I

_{ Total }- I_{ 1 }= I_{ 2 }.25 - .2 = I_{ 2 }= .05 Amperes.The other method would be by using Ohms Law. We know the resistance of R

_{ 2 }= 200 Î©. We also know that the voltage across R_{ 2 }= 10 Volts. Hence:10 Volts / 200 Î© = .05 Amperes.

Either way, our final result is A

_{ 2 }= .05 Amps