At t = 0 the switch S is closed with the capacitor uncharged. If C = 30 micoF, rmf= 50 V, and R = 10 k, what is the potential difference across the resistor as soon as the switch is closed?
assuming they are in series, the entire voltage appears across the resistor immediately because the capacitor has to build up charge on its plates to have a voltage. C= charge / voltage where charge = integral of current over time
To determine the potential difference across the resistor as soon as the switch is closed, we need to use the concepts of RC circuits.
An RC circuit consists of a resistor (R) and a capacitor (C) connected in series to a voltage source. When the switch is closed, the capacitor starts to charge, and the potential difference across the resistor changes over time.
The time constant (τ) of an RC circuit is given by the product of the resistance (R) and the capacitance (C), expressed as τ = R * C. The time constant represents the time it takes for the voltage across the capacitor to reach approximately 63.2% of the final voltage.
In this case, the capacitance (C) is given as 30 micoF, which can be converted to Farads by dividing it by 1,000,000. So, C becomes 30 * 10^(-6) F.
The resistance (R) is given as 10 kΩ, which can be directly used in the calculation.
The time constant (τ) can be calculated by substituting the values: τ = (10,000 Ω) * (30 * 10^(-6) F) = 0.3 seconds.
Since the switch is closed at t = 0, the potential difference across the resistor as soon as the switch is closed is zero.
As time progresses, the potential difference across the resistor will approach the final value given by the product of the current flowing in the circuit (I) and the resistance (R), according to Ohm's Law.
The potential difference across the resistor at a given time (t) after the switch is closed can be calculated using the exponential charging equation:
V = Vf * (1 - e^(-t/τ))
where V is the potential difference across the resistor at time t, Vf is the final potential difference across the resistor, and e is the base of the natural logarithm (approximately 2.71828).
In this case, the final potential difference across the resistor (Vf) is given as 50 V.
Plugging in t = 0 and Vf = 50 V, we can calculate the potential difference across the resistor as soon as the switch is closed:
V = 50 V * (1 - e^(-0/0.3))
Since e^(-0/0.3) = 1, the equation simplifies to:
V = 50 V * (1 - 1) = 50 V * 0 = 0 V
Therefore, the potential difference across the resistor as soon as the switch is closed is zero.