Ohm's law, computing for Ideal sources(electronics) how to find current passing through a point?
I'm new to circuits and stuff I'm just wondering how to find the values for current in an ideal source, dunno how to use ohm's law in this one circuit here is the link to the problem
http://img177.imageshack.us/img177/2474/...
Answer:
You just need to understand that for an ideal current source, the current coming out of it is fixed, come what may. And likewise for an ideal voltage source, the voltage across it cannot change. You also obviously need to know Ohm's law, I = V/R.
In your circuit:
I1 is the current through a 1 ohm resistor with 3V across it, and defined as flowing from the +ve side of the voltage source. By Ohm's law, it must be +3A.
I2 is the current flowing into a 1A current source, i.e. in the opposite direction to the source. It must therefore be –1A.
Vab is the sum of a 2V and a 3V voltage source in series, with their voltages in the same direction and the +ve side at a, so it is +5V.
Other circuits require a bit of algebra to analyse, but you can do this one by simple reasoning as long as you remember the definitions.
This problem is an equation with four unknowns, i1 and i2, Va and Vb. Because you have 4 unknowns you need 4 equations. You'll technically end up with 4 Ohm's Law equations.
Look at each "Loop" of the circuit (notice there are two) as a separate circuit. Solve for i in the one on the left using just variables. With this you can now solve for V, since Ohm's Law is V=I*R.
3(V) = 1(R) * i1
i1 = 3(A)
Once you solved the loop on the left you incorporate it into the loop on the right and you can solve for each of the variables.
Kirchoff's current law says that the sum of all current into and out of a point is zero.
Kirchoff's voltage law says that the sum of voltage drops around a loop is zero.
http://en.wikipedia.org/wiki/kirchhoff%2...
Ohm's law says V = I*R (voltage = current * resistance)
http://en.wikipedia.org/wiki/ohm%27s_law...
i1 is answered using Ohm's law.
i2 is answered using Kirchoff's current law.
v(a+b) is answered using Kirchoff's voltage law.
look the branch which the current source in it =1A is also the branch which the the current ( | )2 passed on it so ( | )2 = -1 A
And the Volt a-b = the voltage source (2v) + the voltage source(3v) and that equal 5V , the voltage drope on the resistance 1 ohm is the voltage source 3V so from ohms law V=IR therefor 3V= ( | )1 * 1 ohm so ( | )1 =3V/1 ohm
( | )1 = 3 A
More Questions and Answers:
Battery lead manufacaturers?
An unspliced grounding electroding conductor shall be permitted to run to _ grounding electrode?
I want to make a design for air bubble detector in moving blood?
Which equipments are used for production of wood pellets?
What Metal is around 256% stronger than steel?
What is meant by direct cold water explained with a diagram?
Which is the formula for calculating the lathe time when the tool moves vertically to the axis of the shaft?
What determines the electrical conductivity of metals? Which metals are better conductors than others?
How green can a household of 3 be, realistically?
Watt - kilo watt usage?
http://img177.imageshack.us/img177/2474/...
Answer:
You just need to understand that for an ideal current source, the current coming out of it is fixed, come what may. And likewise for an ideal voltage source, the voltage across it cannot change. You also obviously need to know Ohm's law, I = V/R.
In your circuit:
I1 is the current through a 1 ohm resistor with 3V across it, and defined as flowing from the +ve side of the voltage source. By Ohm's law, it must be +3A.
I2 is the current flowing into a 1A current source, i.e. in the opposite direction to the source. It must therefore be –1A.
Vab is the sum of a 2V and a 3V voltage source in series, with their voltages in the same direction and the +ve side at a, so it is +5V.
Other circuits require a bit of algebra to analyse, but you can do this one by simple reasoning as long as you remember the definitions.
This problem is an equation with four unknowns, i1 and i2, Va and Vb. Because you have 4 unknowns you need 4 equations. You'll technically end up with 4 Ohm's Law equations.
Look at each "Loop" of the circuit (notice there are two) as a separate circuit. Solve for i in the one on the left using just variables. With this you can now solve for V, since Ohm's Law is V=I*R.
3(V) = 1(R) * i1
i1 = 3(A)
Once you solved the loop on the left you incorporate it into the loop on the right and you can solve for each of the variables.
Kirchoff's current law says that the sum of all current into and out of a point is zero.
Kirchoff's voltage law says that the sum of voltage drops around a loop is zero.
http://en.wikipedia.org/wiki/kirchhoff%2...
Ohm's law says V = I*R (voltage = current * resistance)
http://en.wikipedia.org/wiki/ohm%27s_law...
i1 is answered using Ohm's law.
i2 is answered using Kirchoff's current law.
v(a+b) is answered using Kirchoff's voltage law.
look the branch which the current source in it =1A is also the branch which the the current ( | )2 passed on it so ( | )2 = -1 A
And the Volt a-b = the voltage source (2v) + the voltage source(3v) and that equal 5V , the voltage drope on the resistance 1 ohm is the voltage source 3V so from ohms law V=IR therefor 3V= ( | )1 * 1 ohm so ( | )1 =3V/1 ohm
( | )1 = 3 A
The answers post by the user, for information only, FunQA.com does not guarantee the right.
More Questions and Answers: