Two capacitors 3 farad and 6 farad are connected in series with a 6 volt battery. Which one will have higher potential?
To determine which capacitor will have a higher potential in a series connection, one must first understand how capacitors behave when connected in series.
When capacitors are connected in series, the total capacitance (C_total) can be calculated using the formula:
1/C_total = 1/C1 + 1/C2 + ...
In this case, we have two capacitors, C1 = 3 Farad and C2 = 6 Farad. By substituting these values into the formula, we can find the total capacitance:
1/C_total = 1/3 + 1/6
To simplify the equation, we need to find the least common multiple (LCM) of the denominators, which is 6:
1/C_total = 2/6 + 1/6 = 3/6
Now, we can take the reciprocal of both sides of the equation to find C_total:
C_total = 6/3 = 2 Farad
So, the total capacitance of the series combination is 2 Farad.
Now, when capacitors are connected in series, the voltage across each capacitor depends on its capacitance. The voltage (V) across each capacitor (V1 and V2) can be calculated using the formula:
V = (Q / C),
Where Q is the charge stored in each capacitor, and C is the capacitance.
Since both capacitors are connected in series, they share the same amount of charge (Q), but the voltage across each capacitor will be inversely proportional to their respective capacitances. Therefore, the larger the capacitance, the lower the voltage across it.
In this case, the 3 Farad capacitor has a smaller capacitance than the 6 Farad capacitor, which means it will have a higher voltage across it compared to the 6 Farad capacitor.
So, in the given scenario, the 3 Farad capacitor will have a higher potential compared to the 6 Farad capacitor.