In the given circuit \(C_1 = 2μF, C_2 = 0.2μF, C_3 = 2μF, C_4 = 4μF, C_5 = 2μF, C_6 = 2μF\). The charge stored on capacitor \(C_4\) is _____ \(μC\)
In the given circuit \(C_1 = 2μF, C_2 = 0.2μF, C_3 = 2μF, C_4 = 4μF, C_5 = 2μF, C_6 = 2μF\). The charge stored on capacitor \(C_4\) is _____ \(μC\)
Correct Answer: 4
Solution and Explanation
Step 1: Equivalent Capacitance of the Circuit
The total equivalent capacitance (\( C_{\text{eq}} \)) of the circuit is given as:
\[ C_{\text{eq}} = 0.5 \, \mu\text{F}. \]
Step 2: Total Charge (\( Q \)) in the Circuit
The total charge stored in the circuit can be calculated using the formula:
\[ Q = C_{\text{eq}} \cdot V, \]
where:
- \( C_{\text{eq}} = 0.5 \, \mu\text{F} \) (equivalent capacitance),
- \( V = 10 \, \text{V} \) (applied voltage).
Substitute the values:
\[ Q = 0.5 \cdot 10 = 5 \, \mu\text{C}. \]
Step 3: Charge Redistribution (\( Q' \))
The charge on a specific branch of the circuit is calculated using the charge division formula. For the given branch:
\[ Q' = Q \cdot \frac{C_2}{C_2 + C_6}. \]
Substitute the values:
- \( Q = 5 \, \mu\text{C} \),
- \( C_2 = 0.8 \, \mu\text{F} \),
- \( C_6 = 0.2 \, \mu\text{F} \).
\[ Q' = 5 \cdot \frac{0.8}{0.8 + 0.2}. \]
Simplify:
\[ Q' = 5 \cdot \frac{0.8}{1} = 5 \cdot 0.8 = 4 \, \mu\text{C}. \]
Final Answer:
- \( C_{\text{eq}} = 0.5 \, \mu\text{F} \)
- \( Q = 5 \, \mu\text{C} \)
- \( Q' = 4 \, \mu\text{C} \)
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