# Problem Set 10

```Physics 2220 – Module 10 Homework
01.
What is the potential difference across a 10 mH inductor if the current through the inductor drops from
150 mA to 50 mA in 10 μs? What is the direction of the current?
Use the definition of the potential difference for an inductor.
Δ V = −L
−3
(
−3
dI
50 &times; 10 A − 150 &times; 10
−3
= −(10 &times; 10 H)
−6
dt
10 &times; 10 s
A
)
= 100 Volts
As the current decreases, so does the flux through the center of the inductor. Induced current will
oppose this change, so the induced current will be in the same direction as the original current in order
to try to maintain the original flux. Therefore the potential will increase in the direction of the current.
02.
A 100 mH inductor whose windings have a resistance of 4.0 Ω is connected across a 12 V battery
having an internal resistance of 2.0 Ω. How much energy is stored in the inductor?
The energy stored within a inductor is defined by:
UL=
1
2
LI
2
The current can be determined using Ohm's Law. Note that the internal resistance of the battery and the
resistance of the inductor are in series.
I=
ε battery
Δ V
12 V
=
=
= 2.0 A
R
R inductor + r (4.0 Ω + 2.0 Ω)
Now determine the stored energy:
UL=
03.
1
1
2
−3
2
LI = (100 &times; 10 H) (2.0 A ) = 0.20 J
2
2
An FM radio station broadcasts at a frequency of 100 MHz. What inductance should be paired with a 10
pF capacitor to build a receiver circuit for this station?
Use the definition for resonant frequency to determine the value of the needed inductor.
ω=
L=
1
2
ω C
=
√
1
LC
1
1
=
= 2.5 &times; 10−7 H = 0.25 μ H
2
2
6
2
−12
(2 π f ) C
4 π (100 &times; 10 Hz) (10 &times; 10 F)
04.
What value of resistor R gives the circuit in the figure a time constant of 25 μs?
Use the definition of the time constant for a series RL
circuit:
L
τ=
R equ
The resistance in this equation must be the equivalent
resistance of the two parallel resistors so then the
equivalent resistance and inductor are in series.
L
R equ = τ
1
1
1
=
+ =τ
R equ 500 Ω R
L
−6
1
1
25 &times; 10 s
1
= τ−
=
−
R
L 500 Ω 7.5 &times; 10−3 H 500 Ω
R = 750 Ω
05.
The switch in the figure has been open for a long time. It is closed at t = 0 seconds.
(a)
What is the current through the battery immediately after the switch is closed?
Since the switch has been open for a long
time, it can be assumed that no current is
running through any part of the circuit.
Current cannot change instantly when the
switch is closed:
Δ V L = −L
dI
dt
Current through the inductor was zero initially, so it must be very near zero right when the switch
is closed (t = 0). Thus all the current is going through the 20 Ω resistor that is not in series with
the inductor.
10 V
I= ε =
= 0.5 A
R 20 Ω
(b)
What is the current through the battery after the switch has been closed a long time?
After a long time, the current through the inductor reached a constant value, therefore there is
not change in the current, thus:
Δ V L = −L
dI
=0
dt
An ideal inductor with no resistance can be assumed, so now just add the two resistors in
parallel.
1
1
1
=
+
R equ 20 Ω 20 Ω
→
R equ = 10 Ω
10 V
I= ε =
= 1.0 A
R equ 10 Ω
```