Charging behavior of an RC network — which statements are true as a capacitor charges from a DC source? Assume a simple series R–C connected to a constant DC supply and select the best overall answer.
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Athe voltage across the plates rises exponentially
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Bthe circuit current falls exponentially
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Cthe capacitor charges to the source voltage in 5×RC seconds
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Dall of the above
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Enone of the above
Answer
Correct Answer: all of the above
Explanation
Introduction / Context:Classic RC charging exhibits exponential behavior for both capacitor voltage and loop current. The time constant tau = R * C sets the rate. After about 5 * tau, the capacitor is essentially fully charged for practical purposes. This question checks fundamental time-constant intuition.
Given Data / Assumptions:
- Series R–C driven by an ideal DC source
- Initial capacitor voltage is zero
- No leakage and constant R
Concept / Approach:Standard equations for charging are: Vc(t) = V_s * (1 − e^(−t/tau)) and I(t) = (V_s/R) * e^(−t/tau). Engineers often use 5 * tau as the practical "full-charge" time where Vc ≈ 99%.
Step-by-Step Solution:
tau = R * CVc(t) increases exponentially toward V_sI(t) decreases exponentially toward 0t ≈ 5 * tau gives Vc ≈ 0.993 * V_s, close enough to "charged"Verification / Alternative check:Plotting Vc(t) and I(t) vs time confirms the exponential rise and decay and the 5 * tau rule of thumb.
Why Other Options Are Wrong:
- Options a, b, c are all correct statements; thus "all of the above" is the best choice.
- "None of the above" contradicts established RC theory.
Common Pitfalls:
- Assuming charging is linear; it is exponential.
- Equating 5 * tau with an exact limit; it is a practical guideline.
Final Answer:all of the above