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Physics

1st Edition

Walker

ISBN: 9780133256925

Textbook solutions

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Page 840: Lesson Check

Exercise 32

Step 1

1 of 2

The voltage in an RLC circuit is given at the form $V(t)=Vcdotsin{omega}cdot{t}$. So the voltage in circuit is described with the sinusoidal function.

Result

2 of 2

Sinusoidal function.

Exercise 33

Step 1

1 of 2

The average power in circuit is given by formula:

$$
begin{align*}
P_{avg}&=I^2_{rms}cdot{R}
end{align*}
$$

Result

2 of 2

$$
P_{avg}=I^2_{rms}cdot{R}
$$

Exercise 34

Step 1

1 of 2

The transformer is defined by the equation:

$$
begin{align*}
frac{V_p}{V_s}=frac{N_p}{N_s}
end{align*}
$$

As we have given that $N_s>N_p$, so must be $V_s>V_p$.
This means that the transformer increases the voltage.

Result

2 of 2

Transformer increases the voltage.

Exercise 35

Step 1

1 of 2

The transformer is defined by the equation:

$$
begin{align*}
frac{V_p}{V_s}=frac{N_p}{N_s}=frac{I_s}{I_p}
end{align*}
$$

As we see that the current of transformer secondary side is irreversibly proportional to a voltage, if the voltage is increased, the current is reduced.

Result

2 of 2

if the voltage is increased, the current is reduced.

Exercise 36

Step 1

1 of 2

There is many reason for using the rms values, but the main advantage is that the all instruments measure the average value of voltage and current in one semi period, so for the practical purpose we use rms values.

Result

2 of 2

All instruments measures the average value, so this is the main reason for using rms values.

Exercise 37

Step 1

1 of 2

The transformer is defined by the equation:

$$
begin{align*}
frac{V_s}{V_p}=frac{N_s}{N_p}=frac{I_p}{I_s}
end{align*}
$$

So in our problem:

$$
bold{a)}
$$

$$
begin{align*}
N_p&=N\
N_s&=2cdot{N}\
frac{V_s}{V_p}&=frac{N_s}{N_p}\
frac{V_s}{V_p}&=frac{2cdot{N}}{N}
end{align*}
$$

$$
boxed{frac{V_s}{V_p}=2}
$$

$$
bold{b)}
$$

$$
begin{align*}
N_p&=N\
N_s&=2cdot{N}\
frac{I_s}{I_p}&=frac{N_p}{N_s}\
frac{I_s}{I_p}&=frac{N}{2cdot{N}}
end{align*}
$$

$$
boxed{frac{I_s}{I_p}=0.5}
$$

Result

2 of 2

a) $frac{V_s}{V_p}=2$

b) $frac{I_s}{I_p}=0.5$

Exercise 38

Step 1

1 of 2

The relation between rms and max value is given as:

$$
begin{align*}
I_{max}&=sqrt{2}cdot{I_{rms}}
end{align*}
$$

Let’s substitute and compute:

$$
begin{align*}
I_{max}&=sqrt{2}cdot{2.5text{ A}}
end{align*}
$$

$$
boxed{I_{max}=3.54text{ A}}
$$

Result

2 of 2

$$
I_{max}=3.54text{ A}
$$

Exercise 39

Step 1

1 of 2

The relation between rms and max value is given as:

$$
begin{align*}
V_{rms}&=frac{1}{sqrt{2}}cdot{V_{max}}
end{align*}
$$

Let’s substitute and compute:

$$
begin{align*}
V_{rms}&=frac{1}{sqrt{2}}cdot{5text{ V}}
end{align*}
$$

$$
boxed{V_{rms}=3.54text{ V}}
$$

Result

2 of 2

$$
V_{rms}=3.54text{ V}
$$

Exercise 40

Step 1

1 of 2

The equation to compute the average power dissipated on the resistor is:

$$
begin{align*}
P_{avg}&=I^2_{rms}cdot{R}
end{align*}
$$

Let’s substitute and compute:

$$
begin{align*}
P_{avg}&=(3.2text{ A})^2cdot{180text{ $Omega$}}
end{align*}
$$

$$
boxed{P_{avg}=1843.2text{ W}}
$$

Result

2 of 2

$$
P_{avg}=1843.2text{ W}
$$

Exercise 41

Step 1

1 of 2

The transformer is defined by the equation:

$$
begin{align}
frac{V_s}{V_p}=frac{N_s}{N_p}
end{align}
$$

We have given:

$$
begin{align*}
N_p&=50\
N_s&=125\
V_p&=25text{ V}
end{align*}
$$

$bold{a)}$

As we have more loops at a secondary coil, the transformer increase the voltage so the voltage on a secondary coil is:
$$
boxed{text{Higher than $25text{ V}$.}}
$$

$bold{b)}$

Let’s express the voltage on the secondary coil and substitute:

$$
begin{align*}
V_s&=frac{N_s}{N_p}cdot{V_p}\
V_s&=frac{125}{50}cdot{5text{ V}}
end{align*}
$$

$$
boxed{V_s=62.5text{ V}}
$$

Result

2 of 2

a) Higher then $25text{ V}$.

b) $V_s=62.5text{ V}$

Exercise 42

Step 1

1 of 2

The transformer is defined by the equation:

$$
begin{align}
frac{I_s}{I_p}=frac{N_p}{N_s}
end{align}
$$

We have given:

$$
begin{align*}
N_p&=150\
N_s&=35\
I_p&=2.1text{ A}
end{align*}
$$

$bold{a)}$

As we have more loops at a primary coil, the transformer decrease the voltage, and increase the current at secondary side, so the current on a secondary coil is:
$$
boxed{text{Higher than $2.1text{ A}$.}}
$$

$bold{b)}$

Let’s express the current through the secondary coil and substitute:

$$
begin{align*}
I_s&=frac{N_p}{N_s}cdot{I_p}\
I_s&=frac{150}{35}cdot{2.1text{ A}}
end{align*}
$$

$$
boxed{I_s=9text{ A}}
$$

Result

2 of 2

a) Higher then $2.1text{ A}$

b) $I_s=9text{ A}$

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