Question

In an experiment on charging of an initially uncharged capacitor, an RC circuit is made with the resistance R = 1 kΩ and the capacitor C = 10⁴ uF along with an ideal voltage source of 60 V. If the displacement current through the capacitor (in mA), 5 seconds after the charging has started is 4P then find the value of P.

[Take, √e 5/3]
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Answer 1

In the given RC circuit with R = 1 kΩ and C = 10⁴ µF, the displacement current after 5 seconds of charging is 36 mA. Comparing this to 4P mA, the value of P is 9.

Step-by-step explanation:

The student is asking about the concept of displacement current in an RC circuit specifically while charging a capacitor. In this scenario, we have a resistance (R) of 1 kΩ (1,000 ohms) and a capacitor (C) with a capacitance of 10⁴ µF (10,000 µF) in a circuit with a 60 V voltage source. The displacement current (I) in the capacitor is identified after a duration of 5 seconds since the charging started. According to the RC time constant formula T = RC, where T is in seconds, R is in ohms, and C is in farads, we can determine the time constant for the circuit and consequently calculate the current using the formula for the charging of a capacitor, which is I = (V/R) · e^(-t/T).

The time constant (τ) here is R · C = 1,000 ohms · 10,000 µF = 10 seconds. So at t = 5 seconds (which is 0.5 τ), the current will be I = (60 V / 1,000 ohms) · e^(-5/10) which simplifies to I = 0.06 A · e^(-0.5). After calculation, the current I is approximately 0.036 A or 36 mA. Given in the question, the displacement current is also 4P mA, hence 36 mA = 4P, which means P = 9.