# Question on current

1. Jul 18, 2015

### Lim Y K

The equation for current in series is I1=12=13 as the charges remain constant as it passes through the circuit regardless of the resistances of the resistors. i1 being the current before the resistor, i2 being the current after resistor 1 and i3 being the current after the last (2nd) resistor before going back to the battery. However, I=Q/T and since the resistors have resistance, the current will move slower through it. Wont this affect the current? why the current remains constant throughout the circuit (i1=i2=i3)?? can someone explain to me please?

2. Jul 18, 2015

### phinds

Current is like a bicycle chain. The whole thing moves, and all at the same speed, or it doesn't move at all (if you have an open circuit or a broken chain). That probably doesn't answer you question at the level of fundamental physics, but it IS the way it works.

3. Jul 18, 2015

### Staff: Mentor

The charges don't slow down. Instead the resistance means that we have to push the charges harder to keep them moving at the same speed; that push comes from the voltage, which is higher at one end of the resistor than the other.

4. Jul 18, 2015

### Chandra Prayaga

could you draw a circuit diagram? I am not sure what you mean by "current before" a resistor and after a resistor.

5. Jul 18, 2015

### phinds

Think of a resistor drawn horizontally with the current flowing left to right. The left side is "before" the resistor and the right side is "after" the resistor.

6. Jul 18, 2015

### Staff: Mentor

Ask yourself, what would happen if $I_1 > I_2$. Specifically, what would happen to the charge on the first resistor?

7. Jul 18, 2015

### Lim Y K

there will be more chrage flowing through resistor one?

8. Jul 18, 2015

### Staff: Mentor

More than that. There will be charge accumulating in resistor 1. It will gain more and more net charge the longer the imbalance between the currents persists.

How long do you think any power source can push against the enormous E field that would be created by such a net charge?

9. Jul 18, 2015

### Lim Y K

oh so the reason why there's no charge build-up at any components of circuit is because the current remains the same throughout?

10. Jul 18, 2015

### Staff: Mentor

Yes. If there were 1 A entering a resistor and 0.9 A leaving, then the resistor would be gaining 0.1 C every second. The E fields generated would quickly overwhelm any power source and crush any circuit material. The forces we are talking about are simply enormous.

11. Jul 18, 2015

### Lim Y K

if more resistors are added into a circuit, the current will decrease but it will remain the same throughtout the circuit? or will more voltage be required to keep the current at the same speed?

12. Jul 19, 2015

### Qwertywerty

To your first question - Yes it will - if in series - Resistors only cause potential drop .

Current does not have speed . If you mean that you wish to keep the rate flow of charge constant
( current) , Yes you would have to increase the potential difference applied across the circuit by the battery or cell .

Last edited: Jul 19, 2015
13. Jul 19, 2015

### Drakkith

Staff Emeritus
If the resistors are added in series with other components, and the voltage is kept the same, then the current will decrease.