All Solutions
Page 675: Practice Problems
$EMF=BLv$
a) Now, we can insert the values given in the problem
$$
EMF=0.4times 0.5times 20
$$
which gives that
$$
boxed{EMF=4textrm{V}}
$$
b) Now, can proceed to find the induced current in the circuit by employing Ohm’s law
$$
I=frac{EMF}{R}=frac{4}{6}
$$
so we get that
$$
boxed{I=0.67textrm{A}}
$$
textrm{a) }EMF=4textrm{V}
$$
$$
textrm{b) }I=0.67textrm{A}
$$
$EMF=BLv$
Now, we can insert the values given in the problem
$$
EMF=5times 10^{-5}times 25times 125
$$
which gives that
$$
boxed{EMF=0.16textrm{V}}
$$
EMF=0.16textrm{V}
$$
EMF_{ind}= -BLvsintheta
$$
Ohm’s law (V=I*R)
b) current intensity = 4 A

a) We start from the formula for the effective voltage which says
$$
V_{eff}=frac{sqrt{2}}{2}V_{max}
$$
So after we plug the values
$$
V_{eff}=frac{sqrt{2}}{2}170
$$
$boxed{V_{eff}=120textrm{V}}$
b) The same procedure we repeat to find the effective current
$$
I_{eff}=frac{sqrt{2}}{2}I_{max}
$$
So after we plug the values
$$
I_{eff}=frac{sqrt{2}}{2}0.7
$$
Finally, we have that
$boxed{I_{eff}=49textrm{A}}$
c) At the end, we use Ohm’s law to find the circuit resistance
$$
R=frac{V_{eff}}{I_{eff}}=frac{120}{49}
$$
$$
boxed{R=240Omega}
$$
textrm{a) }V_{eff}=120textrm{V}
$$
$$
textrm{b) }I_{eff}=49textrm{A}
$$
$$
textrm{c) }R=240Omega
$$
$$
V_{RMS}=frac{sqrt{2}}{2}V_{max}
$$
We can express the maximum voltage as follows
$$
V_{max}=sqrt{2}V_{RMS}
$$
So after we plug the values
$$
V_{max}=sqrt{2}times 117
$$
$$
boxed{V_{max}=165textrm{V}}
$$
The same procedure we repeat to find the maximum current
$$
I_{RMS}=frac{sqrt{2}}{2}I_{max}
$$
$$
I_{max}=sqrt{2}I_{RMS}
$$
So after we plug the values
$$
I_{max}=sqrt{2}times 5.5
$$
$$
boxed{I_{max}=7.76textrm{A}}
$$
V_{max}=165textrm{V}
$$
$$
I_{max}=7.76textrm{A}
$$
a) We start from the formula for the effective voltage which says
$$
V_{eff}=frac{sqrt{2}}{2}V_{max}
$$
So after we plug the values
$$
V_{eff}=frac{sqrt{2}}{2}425
$$
$boxed{V_{eff}=300textrm{V}}$
b) Now, we can use Ohm’s law to find the effective current
$$
I_{eff}=frac{V_{eff}}{R}=frac{300}{500}
$$
$$
boxed{I_{eff}=0.6textrm{A}}
$$
textrm{a) }V_{eff}=300textrm{V}
$$
$$
textrm{b) }I_{eff}=0.6textrm{A}
$$
$$
P=Itimes V
$$
then the effective power is given as
$$
P_{eff}=I_{eff}V_{eff}=frac{sqrt{2}}{2}I_{max}times frac{sqrt{2}}{2}V_{max}
$$
This gives that
$$
P_{eff}=frac{P_{max}}{2}
$$
Now, we can express the maximum power as
$$
P_{max}=2P_{eff}=2times 75
$$
Finally,
$$
boxed{P_{max}=150textrm{W}}
$$
P_{max}=150textrm{W}
$$