If there is really a potential drop across resistors

[jaredvert]

How come capacitors receive 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?

 

[Andrew Mason]

How come capacitors receive the full emf? 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?

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

 

[jaredvert]

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

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

 

[Drakkith]

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.

 

[Andrew Mason]

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

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

AM

 

[Drakkith]

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

AM

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

 

[jtbell]

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

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).

 

[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.