Hello All,

I studies Circuit theory almost 2 decades back and hardly applied it in my work. Mostly I tend to work in digital area.

I am getting a lot confused with the concept of voltage & current. If I may explain with an example.

I have 2 series resisters of 2 ohm and 3 ohm connected to a 10 v source. The voltage drop across first resister is 4 V and across second resister is 6 V.

Since the current flowing through 2 ohm and 3 ohm resister is same, how should I interpret what exactly voltage is?. Is it the rate at which current flows through R1 and R2 is different and is related to voltage drop?

Similarly, if I keep increasing the number of resisters in the series then the current comes down. Does that mean that current supply reduces as resistance load on a voltage source increases?

Similarly, if number of series resisters are decreased than current increases. Does that mean that rate at which the current flows across the resister increases as the number of resisters decreases?

Thanks a lot in advance ...

 

Stop using the term "rate".
Rate means a change per unit time.

Go back to Ohm's Law:
The current I through a resistor R is directly proportional to the applied voltage V and inversely proportional to the resistance.

In mathematical terms:
I = V/R

In each of your examples above, simply apply Ohm's Law.

For a fixed voltage V, calculate the current I in a series circuit. Knowing the current I in each resistive element, you can determine the voltage across each element by applying a corollary of Ohm's Law, V = IR.

 

voltage is a force.
current is a flow.
the greater the resistance, the more force it takes for current to flow.

 

Thanks, MrChips & Kermit2. Slowly getting hang of it ...

Instead of passive components, assume I am using a MOS type of transistor. As we know, when transistor is ON, the current flows through the channel.
Now, if I increase the channel width of the transistor keeping eveyrthing else same, my intuition tell me that current increases.

So, what happened in this case, did the resistance decrease due to increase in the channel width?

 

It's all a matter of Ohm's law. Thus if you have a 2 ohms and 3 ohms in series with a 10V source, the current would be I = V/R = 10V / (2Ω+3Ω) = 2A. Then the voltage across the 2Ω resistance is V = I*R = 2A x 2Ω = 4V and across the 3Ω resistor is 2A x 3Ω = 6V. The sum of those is 10V, equal to the supply voltage of course.

From that you should be able to answer the rest of your questions.

 

Thanks, MrChips & Kermit2. Slowly getting hang of it ...

Instead of passive components, assume I am using a MOS type of transistor. As we know, when transistor is ON, the current flows through the channel.
Now, if I increase the channel width of the transistor keeping eveyrthing else same, my intuition tell me that current increases.

So, what happened in this case, did the resistance decrease due to increase in the channel width?

Yes, the resistance decreases.

But when you get to semiconductor devices, a transistor for example, you cannot apply Ohm's Law. That is because the device is non-linear, i.e. the current does not vary proportionately with applied voltage. Another way of looking at it, the resistance is not constant. The resistance changes with applied voltage.

 

It is not wise to jump from, "I don't know the difference between voltage and current." to, "What happens when you change the internal size of a MOS transistor?"

Read this: http://forum.allaboutcircuits.com/threads/ohms-law-for-noobies-or-the-amp-hour-fallacy.69757/

It's very, very, basic concepts for voltage and current.