Electronic – the purpose of the resistor in parallel
resistors
I'm a beginner in electronics, and I have a rather stupid question.
On the circuit below, why is there a need for the second resistor (2 ohm) in parallel to the 2nd lamp?
Best Answer
I'll try to explain:
The 12W lamp needs a certain current, as well as the 60W lamp. If you calculate the currents the lamps need, you should find a difference. So where to put the excess power? Now the 2 Ohm resistor comes into play, it diverts the excess current and that is its task in this circuit.
Understanding this requires Kirchoff's and Ohm's laws:
Kirchhoff's current law : The sum of all currents in a node is equal to zero. This means that all currents flowing into a node flow out again. So in your circuit the current coming from the 60W lamp has to be divided between the 2 Ohm resistor and the 12W lamp.
Kirchhoff's voltage law: The sum of all voltages in a mesh is equal to zero.
Applied to your circuit it means, that the 24 V supply voltage must be distributed to all components. e.g. the 12V at the 60W lamp + the 6V at the 12W lamp + the voltage at the 1.2 Ohm resistor must be equal to the 24 V
Furthermore you have to know that in a parallel circuit all voltages are the same. Applied to your circuit this means that the 2 Ohm resistor and the 12W lamp have the same voltage.
Furthermore, in a series circuit the current is the constant through all the elements, in your circuit the 1.2 Ohm resistor and the 60W lamp are passed through by the same current.
In addition you need to know that the voltage is directly proportional to resistance, which means that the largest resistor will drop the largest voltage.
The resistor is called a pull-down resistor. If it were not there, when the switch is open there would be no defined voltage on the pin the switch is connected to. This is called a "floating" input and generally does not guarantee that the input is in one state or the other. By adding the resistor we know that the input will be low when the switch is open.
Your calculations for both the resistor value and wattage are off.
For the resistor, you show the formula R = V / I, but then calculate using the entire current for all three LEDs (1 A) instead of the current for just one LED.
For the wattage, you are trying to use the formula P = VI, but then you substitute R for the V and use the current for all three resistors (1 A) instead of just one.
Best Answer
I'll try to explain:
The 12W lamp needs a certain current, as well as the 60W lamp. If you calculate the currents the lamps need, you should find a difference. So where to put the excess power? Now the 2 Ohm resistor comes into play, it diverts the excess current and that is its task in this circuit.
Understanding this requires Kirchoff's and Ohm's laws:
Kirchhoff's circuit laws :
Kirchhoff's current law : The sum of all currents in a node is equal to zero. This means that all currents flowing into a node flow out again. So in your circuit the current coming from the 60W lamp has to be divided between the 2 Ohm resistor and the 12W lamp.
Kirchhoff's voltage law: The sum of all voltages in a mesh is equal to zero. Applied to your circuit it means, that the 24 V supply voltage must be distributed to all components. e.g. the 12V at the 60W lamp + the 6V at the 12W lamp + the voltage at the 1.2 Ohm resistor must be equal to the 24 V
Furthermore you have to know that in a parallel circuit all voltages are the same. Applied to your circuit this means that the 2 Ohm resistor and the 12W lamp have the same voltage.
Furthermore, in a series circuit the current is the constant through all the elements, in your circuit the 1.2 Ohm resistor and the 60W lamp are passed through by the same current.
In addition you need to know that the voltage is directly proportional to resistance, which means that the largest resistor will drop the largest voltage.
Power is equal to current times voltage W = V*I
Ohm's law: I = V/R
I think that should be enough to solve the problem.