Would the first Capacitor be twice the voltage as the second in this case, Choice answer d? Then three times as much in the second case? Just wanted to make sense of this. Two capacitors have equal capacitance. If capacitor one is holding twice the amount of charge as the second capacitors, how do the capacitors voltage compare? a. The first capacitor has four times the voltage as the second b. The first capacitor has the equal amount of voltage as the second c. The first capacitor has half the voltage as capacitor two d. The first capacitor has twice the voltage as the second How about given that capacitor one is three times the charge as capacitor two?
Would the first Capacitor be twice the voltage as the second in this case, Choice answer d? Then three times as much in the second case? Just wanted to make sense of this. Two capacitors have equal capacitance. If capacitor one is holding twice the amount of charge as the second capacitors, how do the capacitors voltage compare? a. The first capacitor has four times the voltage as the second b. The first capacitor has the equal amount of voltage as the second c. The first capacitor has half the voltage as capacitor two d. The first capacitor has twice the voltage as the second How about given that capacitor one is three times the charge as capacitor two?
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Chapter1: Units, Trigonometry. And Vectors
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Would the first Capacitor be twice the voltage as the second in this case, Choice answer d? Then three times as much in the second case? Just wanted to make sense of this.
Two capacitors have equal capacitance. If capacitor one is holding twice the amount of charge as the second capacitors, how do the capacitors voltage compare?
a. The first capacitor has four times the voltage as the second
b. The first capacitor has the equal amount of voltage as the second
c. The first capacitor has half the voltage as capacitor two
d. The first capacitor has twice the voltage as the second
How about given that capacitor one is three times the charge as capacitor two?
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