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Unformatted text preview: nt in L1 is given by Equation 23.3 as Irms = Vrms/XL, where
XL = 2π f L1 (Equation 23.4) is the inductive reactance of L1. This current does not depend in
any way on L2 and exists whether or not L2 is present.
The current delivered to the parallel combination is the sum of the currents delivered to each
inductance and is, therefore, greater than either individual current. The current in L2 is given
by Irms = Vrms/XL (Equation 23.3), where XL = 2π f L2 is the inductive reactance of L2
according to Equation 23.4. This current does not depend in any way on L1 and exists
whether or not L1 is present. SOLUTION Using Equation 23.3 to express the current as Irms = Vrms/XL and Equation 23.4
to express the inductive reactance as XL = 2π f L, we have for the current that
I rms = V rms
XL = V rms
2π f L Applying this result to the case where L1 or L2 is connected alone to the generator, we obtain Chapter 23 Problems V
V rms
I 1, rms = rms =
X
2π f L
1444 244441
4L
3 1249 V
V rms
I 2, rms = rms =
X L 2 π f L2
1444 24444
4
3 and L1 alone L2 alone The current delivered to L1 alone is I 1, rms = V rms
2 π f L1 = b 240 V g
c 2 π 2200 Hz 6.0 × 10 −3 H =
h 2 .9 A The current delivered to the parallel combination of L1 and L2 is the sum of that delivered
individually to each inductor and is
I P, rms = I 1, rms + I 2, rms =
= 240 V
2 π 2200 Hz b V rms
2 π f L1 F1
G
H
g6.0 × 10 + F1 + 1 I
2π f L
2π f G L J
L
HK
1
I = 4.8 A
+
J
H 9 .0 × 10 H K
V rms = V rms 2 −3 1 2 −3 13. SSM WWW REASONING Since the capacitor and the inductor are connected in
parallel, the voltage across each of these elements is the same or V L = V C . Using
Equations 23.3 and 23.1, respectively, this becomes I rms X L = I rms X C . Since the currents in
the inductor and capacitor are equal, this relation simplifies to X L = X C . Therefore, we can
find the value of the inductance by equating the expressions (Equations 23.4 and 23.2) for
the inductive reactance and the capacitive reactance, and solving for L.
SOLUTION Since X L = X C , we have 2π f L = 1
2π f C Therefore, the value of the inductance is
L= 1
1
=
= 0.176 H = 176 mH
2
2
2
4 π f C 4 π ( 60.0 Hz) ( 40.0 × 10 –6 F)
2 14. REASONING The inductance L of the inductor determines its inductive reactance XL
according to X L = 2π fL (Equation 23.4), where f is the frequency of the generator. When
the inductor is connected to the terminals of the generator, the rms voltage Vrms of the
generator drives an rms current Irms that depends upon the inductive reactance via 1250 ALTERNATING CURRENT CIRCUITS XL = Vrms
I rms (Equation 23.3). We note that the generator frequency is given in kHz, where 1 kHz = 103 Hz, and the rms current is given in mA, where 1 mA = 10−3 A.
SOLUTION Solving X L = 2π fL (Equation 23.4) for L yields
L= Substituting X L = Vrms
I rms XL (2) 2π f (Equation 23.3) into Equation (1), we find that Vrms Vrms
X L I rms 39 V
=
=
=
= 0.020 H
L=
3
2π f
2π f
2π fI rms 2π ( 7.5 ×10 Hz )( 42 ×10−3 A ) 15. REASONING
a. The inductive reactance XL depends on the frequency f of the current and the inductance
L through the relation XL = 2π f L ( Equation 23.4). This equation can be used directly to
find the frequency of the current.
b. The capacitive reactance XC depends on the frequency f of the current and the capacitance
C through the relation XC = 1/(2π f C ) ( Equation 23.2). By setting XC = XL as specified in
the problem statement, the capacitance can be found.
c. Since the inductive reactance is directly proportional to the frequency, tripling the
frequency triples the inductive reactance.
d. The capacitive reactance is inversely proportional to the frequency, so tripling the
frequency reduces the capacitive reactance by a factor of onethird.
SOLUTION
a. The frequency of the current is f=
b. The capacitance is XL
2.10 ×103 Ω
=
= 1.11×104 Hz
2π L 2π 30.0 ×10−3 H ( ) (23.4) Chapter 23 Problems C= 1
1
=
= 6.83 × 10−9 F
4
3
2π f X C 2π 1.11×10 Hz 2.10 ×10 Ω ( )( ) 1251 (23.2) c. Since XL = 2π f L , tripling the frequency f causes XL to also triple: ( ) X L = 3 2.10 × 103 Ω = 6.30 × 103 Ω d. Since XC = 1/(2π f C ) , tripling the frequency f causes XC to decrease by a factor of 3:
XC = 1
3 ( 2.10 ×103 Ω ) = 7.00 ×102 Ω 16. REASONING AND SOLUTION Equations 23.3 and 23.4 indicate that the rms current in a
single inductance L1 is I1 = V/XL1, where V is the rms voltage and XL1 = 2π fL1. Therefore, c h the current is I 1 = V / 2 π f L1 . Similarly, the current in the second inductor connected c h across the terminals of the generator is I 2 = V / 2 π f L2 . The total current delivered by
the generator is the sum of these two values: I total = I 1 + I 2 = V
V
+
2 π f L1 2 π f L2 But this same total current is delivered to the single inductance L, so it also follows that
I total = V / 2 π f L . Equating the two expressions for Itotal shows that b g V
V
V
=
+
2π f L
2 π f L1 2 π f L2 or 1
1
1
=
+
L
L1 L2 Using this result, we determine the value of L as fo...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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