1. A series RC circuit has a time constant of 1.0 s. The battery has a voltage of 50 V and the maximum current just after closing the switch is 500 mA. The capacitor is initially uncharged. What is the charge on the capacitor 2.0 s after the switch is closed?

The unit of charge is Coulomb.

The charge on the capacitor after 2 second is 0.43 Coulomb.

To finding the voltage while charging a capacitor is given as,

Here , V₀ is the initial potential, R is the resistance and C is the capacitance

It is given that V₀ = 50 V and maximum current I is given as 500 mA.

So,  

Time constant = RC = 1

 So,    

Now, we have  V₀ = 50 V, R = 100 Ω, C = 1/100 F and t = 2 s.

We know that,  

Thus, the charge on the capacitor after 2 s is 0.43 C.

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The charge on the capacitor after 2 s is 0.43 C.

Explanation:

The formula for finding the voltage while charging a capacitor is

Here , V₀ is the initial potential before charging and t is the time at which we have to determine the voltage, R is the resistance and C is the capacitance for the given circuit.

The given problem have given us the values for V₀ = 50 V and maximum current I is given as 500 mA.

Then, resistance can be determined using Ohm's law:

The capacitance is defined as the ratio of charge to the unit voltage.

Here T is the time constant which is given as 1 s and R is found to be 100 ohm, then capacitance will be

So, the values for parameters like V₀ = 50 V, R = 100 Ω, C = 10 mF and t = 2 s.

Then,

V= 43 V.

Then,

Thus, the charge on the capacitor after 2 s is 0.43 C.