Physics

Physics

Problem

A physics student connects an initially uncharged capacitor to an adjustable resistor and a dc power supply. The student begins recording the voltage across the capacitor the instant the circuit is closed. Exactly 3.0 s3.0\,\text{s} after the start of the charging process, the student quickly doubles the resistance without disconnecting the capacitor or power supply and continues the measurement for an additional 12.0 s12.0\,\text{s}. A complete graph of capacitor voltage versus time (starting at t=0t=0 and ending at t=15 st=15\,\text{s}) is obtained. (a) Describe the overall qualitative shape of the voltage-versus-time graph you expect for the entire 15.0 s15.0\,\text{s} interval. Identify any points where the curvature of the graph changes noticeably, and relate those features to events in the experimental procedure. (b) Explain how the graph could be used to estimate the time constant during the first 3.0 s3.0\,\text{s} interval. State any measurements you would take directly from the graph. (c) Explain how the graph could be used to estimate the new time constant after the resistance is doubled. Describe the graphical features you would analyze. (d) Suppose the measured long-time limiting voltage after t=15 st=15\,\text{s} is slightly lower than expected from the battery’s rated voltage. Propose one plausible experimental cause for this discrepancy and describe how it would manifest itself on the graph.
Intuitive explanation

When a capacitor charges through a resistor, the voltage grows exponentially toward the supply voltage with a time constant τ=RC\tau = RC. The key insight is that doubling the resistance midway through charging changes the time constant—and therefore the rate of approach to the final voltage—but not the final voltage itself. You'll see the curve change its curvature at t=3 st = 3\,\text{s}: the first segment rises steeply, then the second segment flattens out as the larger resistance slows the charging. The voltage at t=3 st = 3\,\text{s} becomes the new "starting point" for the second exponential phase.

Intuitive explanation

When a capacitor charges through a resistor, the voltage grows exponentially toward the supply voltage with a time constant τ=RC\tau = RC. The key insight is that doubling the resistance midway through charging changes the time constant—and therefore the rate of approach to the final voltage—but not the final voltage itself. You'll see the curve change its curvature at t=3 st = 3\,\text{s}: the first segment rises steeply, then the second segment flattens out as the larger resistance slows the charging. The voltage at t=3 st = 3\,\text{s} becomes the new "starting point" for the second exponential phase.

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