Please do H I J **Transcription for Educational Website:**---**Figure 3:**Below is an analysis involving an electrical circuit with resistors $ R_1 $ and $ R_2 $, a capacitor $ C $, an electromotive force (emf) $ \varepsilon $, and a switch.### Analysis:**3. In Figure 3 before the switch is closed:**(a) The equation for the charge $ q $ on the capacitor:\[(R_1 + R_2) \frac{dq}{dt} + \frac{q}{C} = \varepsilon - q = C \varepsilon (1 - e^{-\frac{t}{\tau}}) \implies I = \frac{\varepsilon}{R_1 + R_2} e^{-\frac{t}{\tau}}, \ VC = \varepsilon (1 - e^{-\frac{t}{\tau}}); \ V_{Ri} = IR_i = \frac{\varepsilon R_i}{R_1 + R_2} e^{-\frac{t}{\tau}}; \ \tau = (R_1 + R_2) C \ \text{(True, False)}\](b) The displacement current $ I_d $ through the capacitor is equal to the conduction current through the wires i.e.\[I_d = \frac{\varepsilon}{R_1 + R_2}e^{-\frac{t}{\tau}} \ \text{(True, False)}\](c) The time it takes for the voltage across the capacitor to reach half of its maximum value is \[t_{\frac{1}{2}} = (R_1 + R_2) C \ln 2 \ \text{(True, False)}\](d) The power delivered by the battery is \[P_{\varepsilon} = I \varepsilon = \frac{\varepsilon^2}{R_1 + R_2} e^{-\frac{t}{\tau}}\]and energy delivered by the battery at time $ t $ is \[E_{\varepsilon} = \int_0^t P_{\varepsilon} dt = q \varepsilon = \varepsilon^2 C (1 - e^{-\frac{t}{\tau}}) \ \text{(True, False)}\]

Answered: The switch is now closed[the switch has…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Please do H I J

**Transcription for Educational Website:**

---

**Figure 3:**

Below is an analysis involving an electrical circuit with resistors $ R_1 $ and $ R_2 $, a capacitor $ C $, an electromotive force (emf) $ \varepsilon $, and a switch.

### Analysis:

**3. In Figure 3 before the switch is closed:**

(a) The equation for the charge $ q $ on the capacitor:
\[
(R_1 + R_2) \frac{dq}{dt} + \frac{q}{C} = \varepsilon - q = C \varepsilon (1 - e^{-\frac{t}{\tau}}) \implies I = \frac{\varepsilon}{R_1 + R_2} e^{-\frac{t}{\tau}}, \
VC = \varepsilon (1 - e^{-\frac{t}{\tau}}); \ V_{Ri} = IR_i = \frac{\varepsilon R_i}{R_1 + R_2} e^{-\frac{t}{\tau}}; \ \tau = (R_1 + R_2) C \ \text{(True, False)}
\]

(b) The displacement current $ I_d $ through the capacitor is equal to the conduction current through the wires i.e.
\[
I_d = \frac{\varepsilon}{R_1 + R_2}e^{-\frac{t}{\tau}} \ \text{(True, False)}
\]

(c) The time it takes for the voltage across the capacitor to reach half of its maximum value is
\[
t_{\frac{1}{2}} = (R_1 + R_2) C \ln 2 \ \text{(True, False)}
\]

(d) The power delivered by the battery is
\[
P_{\varepsilon} = I \varepsilon = \frac{\varepsilon^2}{R_1 + R_2} e^{-\frac{t}{\tau}}
\]
and energy delivered by the battery at time $ t $ is
\[
E_{\varepsilon} = \int_0^t P_{\varepsilon} dt = q \varepsilon = \varepsilon^2 C (1 - e^{-\frac{t}{\tau}}) \ \text{(True, False)}
\]

Expert Solution

Part H

Let voltage across capacitor be V_C and that across resistor be V_R

Apply KVL in loop 2,

$V_{R} + V_{C} = 0 I R_{2} + \frac{q}{C} = 0 R_{2} \frac{d q}{d t} + \frac{q}{C} = 0 R_{2} \frac{d q}{d t} = - \frac{q}{C} \frac{d q}{q} = - \frac{d t}{R_{2} C} t a k i n g i n t e g r a l o n b o t h s i d e s; \int \frac{d q}{q} = - \int \frac{d t}{R_{2} C} \ln q = - \frac{d t}{R_{2} C} + C \ln q = - \frac{d t}{R_{2} C} + \ln K . . . (\sin c e K i s a c o n s \tan t, \ln K w i l l a l s o b e a c o n s \tan t) T a k i n g a n t i \log o n b o t h s i d e s; q = e^{- \frac{t}{R_{2} C} + \ln K} q = e^{- \frac{t}{R_{2} C}} e^{\ln K} q = K e^{- \frac{t}{R_{2} C}} r e p r e s e n t R_{2} C a s τ; q = K e^{- \frac{t}{τ}}$

Part H

$q = K e^{- \frac{t}{τ}} t a k i n g d e r i v a t i v e w i t h r e s p e c t t o t; I = \frac{d q}{d t} = - \frac{1}{τ} e^{- \frac{t}{τ}} = - \frac{1}{R_{2} C} e^{- \frac{t}{τ}}$

Voltage across the capacitor is;

$V_{c} = \frac{q}{C} = \frac{K e^{- \frac{t}{τ}}}{C} = K' e^{- \frac{t}{τ}} . . . . (\sin c e C i s a f i x e d v a l u e o f c a p a c i t o r, \frac{K}{C} w i l l a l s o b e a f i x e d v a l u e)$

Voltage across the resistance is;

$V_{R} = I R_{2} = - \frac{1}{R_{2} C} e^{- \frac{t}{τ}} \times R_{2} = - \frac{1}{C} e^{- \frac{t}{τ}} S i n c e t h e v a l i e o f c a p a c i t o r i s a f i x e d v a l u e, w e c a n r e w r i t e - \frac{1}{C} a s a c o n s \tan t v a l u e . = K " e^{- \frac{t}{τ}}$

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