Physics 213
Homework 11 Solutions
Spring 2008
30.19.
IDENTIFY:
Apply Kirchhoff’s loop rule to the circuit, sketched below.
i
(
t
) is given by Eq.(30.14).
SET UP:
di
dt
is positive as the current increases from its initial value of zero.
EXECUTE:
±
²
v
R
²
v
L
=
0 , or
²
iR
²
L
di
dt
=
0,
so
i
=
R
1
²
e
²
(
R
/
L
)
t
()
(a)
Initially (
t
= 0),
i
= 0 so
²
L
di
dt
=
0 , and
di
dt
=
L
=
6.00 V
2.50 H
=
2.40 A/s
(b)
²
iR
²
L
di
dt
=
0 (Use this equation rather than Eq.(30.15) since
i
rather than
t
is given.)
Thus
di
dt
=
²
iR
L
=
6.00 V
²
(0.500 A)(8.00
³
)
2.50 H
=
0.800 A/s
(c)
i
=
R
1
²
e
²
(
R
/
L
)
t
=
6.00 V
8.00
³
´
µ
¶
·
¸
¹
1
²
e
²
(8.00
³
/2.50 H)(0.250 s)
=
0.750 A(1
²
e
²
0.800
)
=
0.413 A
(d)
Final steady state means
t
±²
and
di
dt
±
0, so
³
´
iR
=
0. ,
and thus
i
=
R
=
6.00 V
8.00
²
=
0.750 A
EVALUATE:
Our results agree with Fig.30.12 in the textbook. The current is initially zero and increases to its final
value of
/
R
. The slope of the current in the figure, which is
di/dt
, decreases with
t
.
30.61.
IDENTIFY
and
SET UP:
The current grows in the circuit as given by Eq.(30.14). In an
R

L
circuit the full emf initially is
across the inductance and after a long time is totally across the resistance. A solenoid in a circuit is represented as a
resistance in series with an inductance. Apply the loop rule to the circuit; the voltage across a resistance is
V = IR
.
EXECUTE: (a)
In the
R

L
circuit the voltage across the resistor starts at zero and increases to the battery voltage. The
voltage across the solenoid (inductor) starts at the battery voltage and decreases to zero. In the graph, the voltage drops,
so the oscilloscope is across the solenoid.
(b)
At
t
the current in the circuit approaches its final, constant value. The voltage doesn’t go to zero because the
solenoid has some resistance
R
L
. The final voltage across the solenoid is
IR
L
, where
I
is the final current in the circuit.
(c)
The emf of the battery is the initial voltage across the inductor, 50 V. Just after the switch is closed, the current is
zero and there is no voltage drop across any of the resistance in the circuit.
(d)
As
t
,
´
IR
´
IR
L
=
0
±
=50V, and from the graph IR
L
=15V (the final voltage across the inductor), so
IR
=
35 V and
I
=
(35 V)/
R
=
3.5 A
(e)
IR
L
=
15 V, so
R
L
=
(15 V)/(3.5 A)
=
4.3
±
²
V
L
²
iR
=
0, where
V
L
includes the voltage across the resistance of the solenoid.
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 Spring '07
 PERELSTEIN,M
 Current, Inductance, Magnetism, Work, Heat, Resistor, Electrical resistance

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