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So basically, I am inferring the following based off of what I've learned.
When we hook a battery up to a circuit, this creates a potential difference. Say we use a 9V battery. Here is a picture:
From what I know, the horizontal is a equipotential surface, and the vertical has different voltages associated with different points. For example, the top point is 9V, middle 4.5V, bottom 0V.
I have some questions:
(1) Now, let's assume we placed a capacitor on the circuit in the position of the 4.5V dot. Then why do we say the voltage across the capacitor is 9V? Shouldn't the voltage across the capacitor be very tiny, depending on the capacitors size? I.e., the voltage at one plate minus the voltage at the other plate? This certainly wouldn't be 9V.
(2) When we have electrical components on the circuit- why is it that we can say they consume voltage and cause a voltage drop? Isn't this impossible? Isn't it defined that the top is 9V, the bottom is 0V. Therefore, shouldn't it be impossible for this voltage to drop, unless we change the batteries voltage?
(3) If the red line is an equipotential surface, exactly what encourages the current to flow? Is it that the electrons will gather up along this equipotential surface, and eventually end up repelling each other? Thus causing the current to flow?
When we hook a battery up to a circuit, this creates a potential difference. Say we use a 9V battery. Here is a picture:
From what I know, the horizontal is a equipotential surface, and the vertical has different voltages associated with different points. For example, the top point is 9V, middle 4.5V, bottom 0V.
I have some questions:
(1) Now, let's assume we placed a capacitor on the circuit in the position of the 4.5V dot. Then why do we say the voltage across the capacitor is 9V? Shouldn't the voltage across the capacitor be very tiny, depending on the capacitors size? I.e., the voltage at one plate minus the voltage at the other plate? This certainly wouldn't be 9V.
(2) When we have electrical components on the circuit- why is it that we can say they consume voltage and cause a voltage drop? Isn't this impossible? Isn't it defined that the top is 9V, the bottom is 0V. Therefore, shouldn't it be impossible for this voltage to drop, unless we change the batteries voltage?
(3) If the red line is an equipotential surface, exactly what encourages the current to flow? Is it that the electrons will gather up along this equipotential surface, and eventually end up repelling each other? Thus causing the current to flow?