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# System prepended metadata

title: Electronic Circuits Homework 1

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# Electronic Circuits Homework 1
> 劉子雍．108502523
> 資訊工程學系三年級 A 班


## 1.
*(From the left to the right)*
1. $6800\ (\Omega)\ \pm 10 \%$
2. $33\ (\Omega)\ \pm 10 \%$
3. $47000\ (\Omega)\ \pm 5 \%$


## 2.
1. In position 1, $V1 = 0,\ V2 = V_S$.
2. In position 2, $V1 = V_S,\ V2 = 0$.


## 3.
==**ANS**==: (b)


## 4.
### (a)
$R = 10000\ (\Omega)$
$V = 25.0\ (V)$
$I = \frac{V}{R} = \frac{25.0\ (V)}{10000\ (\Omega)} = 0.0025\ (A) = 2.5\ (mA)$
==**ANS**==: $2.5\ (mA)$

### (b)
$R = 2200000\ (\Omega)$
$V = 5.0\ (V)$
$I = \frac{V}{R} = \frac{5.0\ (V)}{2200000\ (\Omega)} \approx 0.00000227\ (A) = 2.27\ (\mu A)$
==**ANS**==: $2.27\ (\mu A)$

### \(c\)
$R = 1800\ (\Omega)$
$V = 15.0\ (V)$
$I = \frac{V}{R} = \frac{15.0\ (V)}{1800\ (\Omega)} \approx 0.00833\ (A) = 8.33\ (mA)$
==**ANS**==: $8.33\ (mA)$


## 5.
### (a)
$R = \frac{V}{I} = \frac{8\ (V)}{2\ (A)} = 4\ (\Omega)$
==**ANS**==: $4\ (\Omega)$

### (b)
$R = \frac{V}{I} = \frac{12\ (V)}{4\ (mA)} = \frac{12\ (V)}{4 \times 10^{-3}\ (A)} = 3 \times 10^3\ (\Omega) = 3\ (k\Omega)$
==**ANS**==: $3\ (k\Omega)$

### \(c\)
$R = \frac{V}{I} = \frac{30\ (V)}{150\ (\mu A)} = \frac{30\ (V)}{150 \times 10^{-6}\ (A)} = 0.2 \times 10^6\ (\Omega) = 200\ (k\Omega)$
==**ANS**==: $200\ (k\Omega)$


## 6.
$P = I^2 R = (100\ (\mu A))^2 \cdot (10\ (k\Omega)) = (100 \times 10^{-6}\ (A))^2 \cdot (10 \times 10^3\ (\Omega)) \\
\quad = 10^{-8} \times 10^{4}\ (W) = 10^{-4}\ (W) = 100\ (\mu W)$
==**ANS**==: $100\ (\mu W)$


## 7.
$P = \frac{V^2}{R} = \frac{(60\ (V))^2}{620\ (\Omega)} = \frac{3600\ (V^2)}{620\ (\Omega)} \approx 5.81\ (W)$
==**ANS**==: $5.81\ (W)$


## 8.
$R_1 = \frac{V_1}{I_1} = \frac{0.5\ (V)}{1\ (A)} = 0.5\ (\Omega)$
$R_2 = \frac{V_2}{I_2} = \frac{1\ (V)}{1\ (A)} = 1\ (\Omega)$
$R_3 = \frac{V_3}{I_3} = \frac{2\ (V)}{1\ (A)} = 2\ (\Omega)$
==**ANS**==: $R_1 = 0.5\ (\Omega),\ R_2 = 1\ (\Omega),\ R_3 = 2\ (\Omega)$


## 9.
$W = 300\ (W) \times 30\ (days) = 0.3\ (kW) \times 30 \cdot 24\ (hours) = 216\ (kWh)$
==**ANS**==: $216\ (kWh)$


## 10.
$I_{minimum} = \frac{120 - 100\ (V)}{8\ (\Omega)} = \frac{20\ (V)}{8\ (\Omega)} = 2.5\ (A)$
==**ANS**==: $2.5\ (A)$

