Unformatted text preview: n arrangement B the two resistors in series have a combined resistance of 2R. Each
series combination is in parallel, so that the reciprocal of the equivalent resistance is
1
1
1
1
=
+
= , or Req, B = R . Following the technique outlined in Section 20.8, we
Req, B 2 R 2 R R
2
also find that Req, C = 3 R and Req, A = 5 R .
4 13. (d) The internal resistance r of the battery and the resistance R are in series, so that the
V
current from the battery can be calculated via Ohm’s law as I =
. The voltage between
R+r
VR
. This result can be solved to show that R = r.
the terminals is, then, 1 V = IR =
2
R+r
14. (b) Kirchhoff’s junction rule states that the sum of the magnitudes of the currents directed
into a junction equals the sum of the magnitudes of the currents directed out of the junction.
Here, this rule implies that I1 + I3 = I 2 .
15. (d) Once the current in the resistor is drawn, the markings of the plus and minus sign are
predetermined. This is because conventional current always flows from a higher toward a
lower potential. Thus, since the current I4 is directed from right to left, the right side of R4
must be marked plus and the left side minus.
16. (b) Kirchhoff’s loop rule states that, around any closed loop, the sum of the potential drops
equals the sum of the potential rises.
17. (a) The ammeter must be connected so that the current that flows through the resistor R2
also flows through the ammeter. The voltmeter must be connected across the resistor R2. Chapter 20 Answers to Focus on Concepts Questions 1055 18. (e) The two capacitors in series have an equivalent capacitance CS that can be determined
1
11
from
= + , so that CS = 1 C . This capacitance is in parallel with a capacitance C, so
2
CS C C
3
that the total equivalent capacitance is Ceq = 1 C + C = 2 C .
2 19. (c) The time constant is given by the product of the resistance and the capacitance.
Therefore, when the resistance is reduced to onethird of its initial value, the capacitance
must be tripled, in order that the time constant remains unchanged.
20. 1.8 s 1056 ELECTRIC CIRCUITS CHAPTER 20 ELECTRIC CIRCUITS
PROBLEMS
______________________________________________________________________________
1. REASONING The current I is defined in Equation 20.1 as the amount of charge ∆q per
unit of time ∆t that flows in a wire. Therefore, the amount of charge is the product of the
current and the time interval. The number of electrons is equal to the charge that flows
divided by the magnitude of the charge on an electron.
SOLUTION
a. The amount of charge that flows is ( ) ∆q = I ∆t = (18 A ) 2.0 × 10−3 s = 3.6 × 10−2 C b. The number of electrons N is equal to the amount of charge divided by e, the magnitude
of the charge on an electron.
3.6 × 10−2 C
∆q
17
N=
=
= 2.3 × 10
−19
e 1.60 × 10 C
______________________________________________________________________________ 2. REASONING We are given the average current I and its duration ∆t. We will employ
∆q
(Equation 20.1) to determine the amount ∆q of charge delivered to the ground by
I=
∆t
the lightning flash.
SOLUTION Solving I = ∆q
(Equation 20.1) for ∆q, we obtain
∆t ( ) ∆q = I ∆t = 1.26 × 103 A ( 0.138 s ) = 174 C ______________________________________________________________________________
3. SSM REASONING Electric current is the amount of charge flowing per unit time (see
Equation 20.1). Thus, the amount of charge is the current times the time. Furthermore, the
potential difference is the difference in electric potential energy per unit charge (see
Equation 19.4), so that, once the amount of charge has been calculated, we can determine
the energy by multiplying the potential difference by the charge.
SOLUTION
a. According to Equation 20.1, the current I is the amount of charge ∆q divided by the time
∆t , or I = ∆q/∆t . Therefore, the additional charge that passes through the machine in normal
mode versus standby mode is Chapter 20 Problems qadditional = I normal ∆t
1 24
43 ∆q in normal mode − I standby ∆t
1 24
43 ( 1057 ) = I normal − I standby ∆t ∆q in standby mode = ( 0.110 A − 0.067 A )( 60.0 s ) = 2.6 C
b. According to Equation 19.4, the potential difference ∆V is the difference ∆(EPE) in the
electric potential energy divided by the charge qadditional, or ∆V = ∆(EPE)/qadditional. As a
result, the additional energy used in the normal mode as compared to the standby mode is
∆ ( EPE ) = qadditional ∆V = ( 2.6 C ) (120 V ) = 310 J 4. REASONING
a. According to Ohm’s law, the current is equal to the voltage between the cell walls
divided by the resistance.
+ b. The number of Na ions that flow through the cell wall is the total charge that flows
divided by the charge of each ion. The total charge is equal to the current multiplied by the
time. SOLUTION
a. The current is I= V 75 × 10−3 V
=
= 1.5 ×10−11 A
R 5.0 ×109 Ω (20.2) + b. The number of Na ions is the total charge ∆q that flows divided by the charge +e on
each ion, or ∆q / e . The charge is t...
View
Full Document
 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

Click to edit the document details