
If the total charge stored in capacitors is equal to 100μc, then find the value of x. (10V)
Understanding the Problem
We are given three capacitors connected in parallel, with capacitances 5 μF, x μF, and 2 μF. The total charge stored is 100 μC when a potential difference of 10 V is applied. We need to find the value of x.
Solution
1. Use the Formula for Total Capacitance in Parallel:
When capacitors are connected in parallel, the total capacitance \( C_{\text{eq}} \) is the sum of the individual capacitances:
\( C_{\text{eq}} = C_1 + C_2 + C_3 \)
Given \( C_1 = 5 \, \mu\text{F} \), \( C_2 = x \, \mu\text{F} \), and \( C_3 = 2 \, \mu\text{F} \), we have:
\( C_{\text{eq}} = 5 + x + 2 = (7 + x) \, \mu\text{F} \)
2. Use the Formula for the Charge Stored in a Capacitor:
The charge stored on the capacitors is given by the formula:
\( Q = C_{\text{eq}} V \)
From the problem, we know that the total charge \( Q = 100 \, \mu\text{C} \) and \( V = 10 \, \text{V} \). Substituting the values into the equation:
\( 100 = (7 + x) \times 10 \)
3. Solve for x:
\( 100 = 70 + 10x \)
\( 10x = 100 - 70 = 30 \)
\( x = \frac{30}{10} = 3 \)
Corrected Final Answer
Thus, the value of \( x \) is \( 3 \).
Note: The original calculation had an error in summing the constant terms.
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