Physics, 04.04.2020 05:59 gabby1274

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?

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Answer from: shod74

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,

$V=V_{0}(1-e^{-\frac{t}{RC} } )$

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, $R=\frac{V}{I}=\frac{50}{500*10^{-3} } =100ohm$

Time constant = RC = 1

So, $C=\frac{1}{R} =\frac{1}{100}F$

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

$V=50(1-e^{-\frac{2}{100*\frac{1}{100} } } )=50(1-e^{-2} )=50(1-0.1353)=43V$

We know that, $Q=CV=\frac{1}{100}*43=0.43C$

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

Learn more about series RC circuit here:

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Answer from: libertycooper

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

Explanation:

The formula for finding the voltage while charging a capacitor is

$V = V_{0}(1-e^{\frac{-t}{RC} })$

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: $R = \frac{V}{I} =\frac{50}{500*10^{-3} } = 100 Ohm$

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

$C = \frac{Q}{V} = \frac{IT}{IR} = \frac{T}{R}$

Here T is the time constant which is given as 1 s and R is found to be 100 ohm, then capacitance will be $\frac{1}{100} = 10 mF$

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

Then, $V =50*(1-e^{\frac{-2}{100*10*10^{-3} } }) =50*(1-e^{-2}) = 50*(1-0.1353)$

V= 43 V.

Then, $Q = CV = 10*10^{-3} * 43 V = 0.43 C$

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

Answer from: Quest

a. are

Answer from: Quest

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6 $\frac{m}{s^{2} }$

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

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1. A series RC circuit has a time constant of 1.0 s. The battery has a voltage of 50 V and the maxim...

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