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A capacitor of capacitance 500×10^(-6)F is charged at a steady rate of 100×10^(-6)C/s . The potential difference across the capacitor will be 10 V in time (A) 25 s (B) 50 s (C) 75 s (D) 100 s?
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A capacitor of capacitance 500×10^(-6)F is charged at a steady rate of...
Calculating the time taken for the potential difference across the capacitor to reach 10V:
- Given data:
- Capacitance (C) = 500×10^(-6)F
- Rate of charging (dQ/dt) = 100×10^(-6)C/s
- Potential difference (V) = 10V
- The formula relating capacitance, charge, and potential difference is: Q = CV
- Rearranging the formula, we get: V = Q/C
- Substituting the given values: 10V = (100×10^(-6)C/s) * t / 500×10^(-6)F
- Simplifying the equation, we get: t = 500s
Therefore, the potential difference across the capacitor will be 10V in 500 seconds.

Answer: (C) 500 s
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A capacitor of capacitance 500×10^(-6)F is charged at a steady rate of 100×10^(-6)C/s . The potential difference across the capacitor will be 10 V in time (A) 25 s (B) 50 s (C) 75 s (D) 100 s?
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A capacitor of capacitance 500×10^(-6)F is charged at a steady rate of 100×10^(-6)C/s . The potential difference across the capacitor will be 10 V in time (A) 25 s (B) 50 s (C) 75 s (D) 100 s? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A capacitor of capacitance 500×10^(-6)F is charged at a steady rate of 100×10^(-6)C/s . The potential difference across the capacitor will be 10 V in time (A) 25 s (B) 50 s (C) 75 s (D) 100 s? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A capacitor of capacitance 500×10^(-6)F is charged at a steady rate of 100×10^(-6)C/s . The potential difference across the capacitor will be 10 V in time (A) 25 s (B) 50 s (C) 75 s (D) 100 s?.
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