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In series combination of capacitors, potential drops across the individual capacitors are equal to
- a)inverse ratio of charges stored.
- b)direct ratio of capacitors
- c)direct ratio of charges stored
- d)inverse ratio of capacitors
Correct answer is option 'D'. Can you explain this answer?
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In series combination of capacitors, potential drops across the indivi...
Explanation:
When capacitors are connected in series, they have equal charge but the potential difference across them is given by
When capacitors are connected in series, they have equal charge but the potential difference across them is given by
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In series combination of capacitors, potential drops across the indivi...
In a series combination of capacitors, the capacitors are connected end-to-end, with the positive terminal of one capacitor connected to the negative terminal of the next capacitor. This creates a single path for the flow of current.
When capacitors are connected in series, the total capacitance (C_total) of the combination can be calculated using the formula:
1/C_total = 1/C1 + 1/C2 + 1/C3 + ...
where C1, C2, C3, ... are the capacitances of the individual capacitors.
In a series combination, the potential difference (V) across the combination is equal to the sum of the potential differences across each capacitor. In other words, the potential drop across each capacitor is the same.
Let's consider two capacitors, C1 and C2, connected in series. The charge stored on each capacitor, Q1 and Q2 respectively, can be calculated using the formula:
Q1 = C1 * V1
Q2 = C2 * V2
where V1 and V2 are the potential drops across C1 and C2 respectively.
Now, we can rearrange these equations to express V1 and V2 in terms of Q1 and Q2:
V1 = Q1 / C1
V2 = Q2 / C2
Since the potential drop across each capacitor is the same, we can equate V1 and V2:
Q1 / C1 = Q2 / C2
Now, let's consider the ratio of the capacitances:
C1 / C2 = Q1 / Q2
Comparing this equation with the equation V1 = Q1 / C1 and V2 = Q2 / C2, we can see that the potential drops across the individual capacitors are inversely proportional to their capacitances.
When capacitors are connected in series, the total capacitance (C_total) of the combination can be calculated using the formula:
1/C_total = 1/C1 + 1/C2 + 1/C3 + ...
where C1, C2, C3, ... are the capacitances of the individual capacitors.
In a series combination, the potential difference (V) across the combination is equal to the sum of the potential differences across each capacitor. In other words, the potential drop across each capacitor is the same.
Let's consider two capacitors, C1 and C2, connected in series. The charge stored on each capacitor, Q1 and Q2 respectively, can be calculated using the formula:
Q1 = C1 * V1
Q2 = C2 * V2
where V1 and V2 are the potential drops across C1 and C2 respectively.
Now, we can rearrange these equations to express V1 and V2 in terms of Q1 and Q2:
V1 = Q1 / C1
V2 = Q2 / C2
Since the potential drop across each capacitor is the same, we can equate V1 and V2:
Q1 / C1 = Q2 / C2
Now, let's consider the ratio of the capacitances:
C1 / C2 = Q1 / Q2
Comparing this equation with the equation V1 = Q1 / C1 and V2 = Q2 / C2, we can see that the potential drops across the individual capacitors are inversely proportional to their capacitances.
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In series combination of capacitors, potential drops across the individual capacitors are equal toa)inverse ratio of charges stored.b)direct ratio of capacitorsc)direct ratio of charges storedd)inverse ratio of capacitorsCorrect answer is option 'D'. Can you explain this answer?
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