# If there is really a potential drop across resistors

1. May 10, 2014

### jaredvert

How come capacitors recieve the full emf In a circuit with a resistor in it???? I plugged in 100000 into -t in the rc transient and it gives me the emf. Why??? I thought you actually "lost" potential through resistors?

Last edited: May 10, 2014
2. May 10, 2014

### Andrew Mason

If you are asking why (in a circuit consisting of a DC source and capacitor) the voltage across the capacitor is the full voltage of the DC source, the answer is: because the capacitor is fully charged. V = Q/C. Once the voltage across the capacitor is equal and opposite to the applied voltage, no current flows.

AM

3. May 10, 2014

### jaredvert

I'm sorry I thought I implied that this was an rc circuit and since there is a potential drop across resistors, im asking why a capacitor would recieve the full emf without losing any voltage to the resistors

4. May 11, 2014

### Drakkith

Staff Emeritus
For this DC circuit you can consider the capacitor to be a variable resistor. Initially there is zero resistance and only the resistors have potential drop across them. But, as the capacitor charges the buildup of charges acts like an increase in resistance and gradually the voltage drop across the capacitor increases. Once the capacitor is fully charged it acts like a resistor with infinite resistance, which means that all of the voltage drop is across the capacitor and none of it is across the resistor.

Try this. Draw a simple circuit with two resistors, R1 and R2, and a battery applying 10 volts. R1 has a resistance of 100 ohms. R2 initially has a resistance of 0 ohms. Find the voltage drop across each resistor. Then increase R2 to 100 ohms, 1000 ohms, and 100,000 ohms and find the voltage drop across each resistor. You will find that the voltage drop across R2 starts at 0 and increases to nearly 12 volts. The capacitor acts in a similar manner.

5. May 11, 2014

### Andrew Mason

The voltage drop across the resistor is IR. If I=0, which occurs when the capacitor is fully charged, IR=0.

AM

6. May 11, 2014

### Drakkith

Staff Emeritus
IR is the voltage drop only for the resistors, correct?

7. May 11, 2014

### Staff: Mentor

Correct. And the voltage drop across the capacitor is Q/C where Q is the charge on the capacitor at the moment.

Assuming this is a simple loop circuit with a battery, resistor and capacitor in series, as you go around the complete circuit the sum of the voltage gains and drops has to be zero at all times. If the battery voltage is E, then E - IR - Q/C = 0. E is constant. As the capacitor charges up, Q increases, so IR decreases. When the capacitor is fully charged, IR = 0 and E = Q/C (or Q = CE).

8. May 11, 2014

### Physics_UG

Capacitors behave like open circuits to DC. Therefore there is no current flowing (because of the open circuit capacitor). No current means no voltage drop across the resistor.