Question

Question: In the circuit shown in Fig. E26.49, C = 5.90 mF, Ԑ = 28.0 V, and the emf has negligible resistance. Initially, the capacitor is uncharged and the switch S is in position 1. The switch is then moved to position 2 so that the capacitor begins to charge. (a) What will be the charge on the capacitor a long time after S is moved to position 2? (b) After S has been in position 2 for 3.00 ms, the charge on the capacitor is measured to be 110 mC What is the value of the resistance R? (c) How long after S is moved to position 2 will the charge on the capacitor be equal to 99.0% of the final value found in part (a)?

Short answer

Answer:

(a) the charge on the capacitor a long time after S is moved to position 2 is

(b) the value of the resistance R is 463 Ω

(c) the charge on the capacitor is equal to 99.0% of the final value found in part (a) is 12.6 ms.

Step-by-step solution

Concept

A capacitor is a device that stores electrical energy in an electric field. It is a passive electronic component with two terminals. The effect of a capacitor is known as capacitance.

Given data

Capacitance C = 5.90 mF

Ԑ = 28.0V

Emf has negligible resistance.

Calculate the charge on the capacitor for different positions of switches.

Calculation for charge

(a)

for the charge on the capacitor, a long time after S is moved to position 2.

The capacitor is charged by the battery in a series resistor when the switch is in position 2.

Charge on the capacitor, a long time after S moves,

$$\begin{gathered} Q_f=\left(\text{28.0}V\right)\left(\text{5.90}\times {\text{10}}^{\text{- 6}}F\right) \\ {Q}_f=\text{165.2}\times {\text{10}}^{\text{- 6}}C\left(\frac{{\text{10}}^{\text{- 6}}C}{\text{1}\mu C}\right) \\ {Q}_f=\text{165.2}\mu C \end{gathered}$$

Calculation of resistance

(b)

forthe value of the resistance R.

(1)

Substituting the given data in the above equation we get,

Thus, the value of the resistance R is

Calculation of time constant

(c)

for the charge on the capacitor to be equal to 99.0% of the final value found in part (a).

Time is given as,

$t=-RC\times ln\left(1-\frac{q}{Q_f}\right)$ (2)

for the capacitor value = 99.0% ,the value of part (a) is,

$\frac{q}{Q_f}=\text{0.99}$

Substitute these values in equation (2), and we get

$$\begin{gathered} t=-\left(\text{463}\Omega \right)\left(\text{5.90}\times {\text{10}}^{\text{- 6}}F\right)\times ln\left(1-0.99\right) \\ t=\text{0.0126}s\left(\frac{1ms}{{10}^{-3}s}\right) \\ t=\text{12.6}ms \end{gathered}$$

thus, the charge on the capacitor is equal to 99.0% of the final value found in part (a) is 12.6 ms