Answered: Answer Part D. please

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN: 9780133923605

Author: Robert L. Boylestad

Publisher: PEARSON

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Answer Part D. please

### Circuit Analysis Problem

This problem involves analyzing the behavior of a capacitor in an electrical circuit. The diagram shows a series circuit containing a 100V DC power source, a 10Ω resistor, and a 50mF capacitor. A switch is present, and it can be toggled between positions "a" and "b".

#### Questions:

a) **How long does it take the capacitor to reach 90% of its maximum charge?**
- This is when the switch is put in position "a" at t=0.

b) **What is the current at this time?**

c) **What is $V_R$ (Voltage across the resistor) at this time?**

d) **What is the rate $U_C$ at which energy is being stored in the capacitor at this time?**

Consider the following for solving the problem:
- When the switch is in position "a," the charging process for the capacitor begins.
- Use the formula for the charging of a capacitor in an RC circuit:
\[
V(t) = V_{\text{max}}(1 - e^{-t/RC})
\]
where $V_{\text{max}}$ is the maximum voltage (100V), $R$ is the resistance, $C$ is the capacitance, and $t$ is time.
- The current $I(t)$ can be found using $I(t) = \frac{V_{\text{max}}}{R} e^{-t/RC}$.
- The voltage across the resistor at any time $t$ is $V_R(t) = I(t) \times R$.
- The energy stored in the capacitor as a function of time can be calculated using $U_C(t) = \frac{1}{2} C V(t)^2$.

Expert Solution

Step 1

According to question:

$C = 50 mF R = 20 Ω ε = 100 V ∵ q = ε C (1 - e^{- \frac{t}{R C}}) and q_{m a x} = ε C \Rightarrow q_{m a x} = 100 \times 50 \times 10^{- 3} = 5 C Now q = 0.9 q_{m a x} = 4.5 C ∴ 4.5 = 5 (1 - e^{- \frac{t}{20 \times 50 \times 10^{- 3}}}) \Rightarrow \frac{4.5}{5} = 1 - e^{- \frac{t}{20 \times 50 \times 10^{- 3}}} \Rightarrow e^{- \frac{t}{20 \times 50 \times 10^{- 3}}} = 0.1 \Rightarrow - \frac{t}{20 \times 50 \times 10^{- 3}} = \ln (0.1) = - 2.302 \Rightarrow t = 2.302 sec ∵ E = \frac{1}{2} C V_{C}^{2} ∴ \frac{d E}{d t} = \frac{2 C V_{C}}{2} \frac{d V_{C}}{d t} = C V_{C} \frac{d V_{C}}{d t} Now \Rightarrow V_{C} (t) = 100 (1 - e^{- \frac{t}{20 \times 50 \times 10^{- 3}}}) \Rightarrow V_{C} (t) = 100 (1 - e^{- t}) V \Rightarrow \frac{d V_{C}}{d t} = 100 e^{- t} ∴ V_{C} (t = 2.302 s e c) = 90 V ∴ {[\frac{d V_{C}}{d t}]}_{t = 2.302 sec} = 10 V/s ∴ {[\frac{d E}{d t}]}_{t = 2.302 sec} = C V_{C} \frac{d V_{C}}{d t} = 50 \times 10^{- 3} \times 90 \times 10 = 45 J/s$

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