Download 13 Electric Circuits

13 Electric Circuits
Answers to Questions
Q1
Q2
Q3
Q4
Q5
Q6
Q7
Q8
Q9
Q10
Q11
Q12
Q13
Q14
Q15
Q16
Q17
Q18
Q19
Q20
Q21
Q22
Q23
Q24
Q25
The bulb will light in arrangement A in which the filament of the bulb is connected to the two sides of the
battery for a closed circuit. In B there is no voltage across the filament and thus no current in the filament.
For A, a lead from the contact in the base of the bulb to the positive terminal of the battery will complete the
circuit. For B, a connection from sides of the screw base of the bulb to the positive terminal of the battery is
necessary.
No. Current (i.e., charge) is conserved. The amount of current entering a point in a circuit is the same as the
amount leaving that point.
No. Electric current is moving electric charge.
The bulb will not light since (dry) wood is a very poor conductor. The resistance will be so high that virtually no
current is in the lamp circuit.
No. Connecting A and B will provide a short circuit for the battery that will damage it while allowing virtually no
current in the bulb.
Diagram B will allow the light bulb to light since there is a closed circuit providing current from the battery
through the bulb. Whether the switch is open or closed is immaterial here since it is in parallel with another
conductor. In diagram A no potential difference is in the closed circuit.
No. The metal clamp will provide a conducting path across the battery causing the battery to discharge. If we
want to use a clamp we can put insulating tape between one of its jaws and the electrical connection.
We had better pay attention to the high voltage warning. The other is a practical joke. The danger to the
body, and even to life, comes from electrical current in the body, which could occur if you accidentally make
contact with a large potential difference across parts of your body. The effect of a high resistance is to limit
current in a circuit if a voltage source is present; it is not dangerous at all.
Decrease. For a conductor obeying Ohm's Law, decreasing the potential difference across a conductor
(resistor), sdecreases the current proportionally.
A good voltmeter is a high resistance device requiring only a small current to actuate it. A "dead" battery has
a high internal resistance so that it is unable to provide sufficient current to light a bulb, but it can still provide a
small current to the voltmeter with a resulting small voltage drop across the internal resistance.
The voltage across the terminals of a battery under load is less than when there is no load because of the
voltage drop across the internal resistance. This is the standard way that auto service station technicians
check the condition of a battery. A large difference between the voltage readings with and without a
resistance across the terminals indicates high internal resistance and a nearly discharged battery.
a. The current is the same in each, since it is a series circuit.
b. The voltage difference is greater across R2. According to Ohm's Law, V = I x R, so for the same current,
the larger the resistance the greater the potential difference
R3 has the greatest current since the current in it is the sum of the currents in R1 and R2.
R3 has the largest voltage difference across it because it has the largest current and the largest resistance.
The current in R3 will decrease because disconnecting R2 will result in a larger total resistance in the circuit.
A property of resistors in series is that they act as if they were a single resistor with a resistance value
equal to the sum of the component resistors. Because of charge conservation the number of charges
flowing through a circuit does not vary.
The correct statement is a. A voltmeter is a high-resistance device connected in parallel with whatever circuit
element is desired to measure the voltage across.
The answer is c. An ammeter is a low-resistance device and is to be placed in series in the circuit, just as a
flow-meter is placed in a fluid circuit.
A voltmeter normally has a high-resistance compared to an ammeter, so it can be placed across two points of
a circuit at different potentials. An ammeter is a low-resistance device placed in series with the rest of the
circuit.
Electric power is the rate of use of electric energy, or in other words, the rate of transfer of electric energy.
P = I2R. The power dissipated will quadruple. The power dissipated is proportional to the square of the
current.
Yes, as can be seen by the equation P = I2R. Power is proportional to resistance.
A pump can increase the potential energy of the water behind the dam by pushing the water higher above
ground level.
No. A battery is based on a chemical reaction that creates a potential difference as a side effect. When the
reaction is complete there will be no more potential difference being generated.
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Q26
Q27
Q28
Q29
Q30
No. Since the current in an a.c. circuit is continuously changing directions, the voltmeter is made to respond to
the magnitude of the effective voltage and no polarity has to be observed nor is any meaningful.
A coffeemaker has a much higher power rating than the shaver or the television and will be the worst offender
on an already loaded circuit. A typical coffeemaker is rated at 1000 watts on a 120-volt line, so that it has a
resistance of 14.4 ohms and will draw 8.3 amperes from the line.
No. To protect a circuit the fuse must be placed in series with the other elements. A fuse across an element
just provides a conducting path around the element and will probably blow without providing any protection.
Should one appliance fail, the whole system would be down. One would be forced to test each appliance
until the culprit was discovered.
The bimetallic strip consists of two fused metals each with a different heat expansion rate. As the heat
inside the appliance rises, the strip bends because this difference in expansion rates causes one of the
metals to expand faster than the other.
Answers to Exercises
E1
6A
E2
150 C
E3
0.25 A
E4
27 V
E5
200 ohms
E6
a. 0.10 A
b. 2 V
E7
a. 0.12 A
b. 4.8 V for 40 ohm, 7.2 V for 60 ohm
E8
a. 0.3 A
b. 2.7 V
E9
a. 0.1 A
b. Yes
c. 2 V
E10
4 ohms
E11
1 ohm
E12
a. 8 ohms
b. 1.5 A
c. 0.5 A
E13
13.5 Watts
E14
a. 0.1 A
b. 0.3 W
E15
a. 0.545 A
b. 201.7 ohms
E16
a. 770 W
b. 15.7 ohms
E17
25A
Answers to Synthesis Problems
SP1
a. 4 ohms
b. 0.125 A
c. 0.0833 A
d. 125 mW
e. The current flowing through the eight ohm resistor is greater than that flowing through the six ohm
resistor because the former is connected in series while the later is connected in parallel.
SP2
a. 0.15 A
b. 0.05 A
c. No. All bulbs are connected in parallel.
SP3
a. 0.043 A
b. 0.86 V
c. 0.386 W
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SP4
SP5
d.
a.
b.
c.
d.
a.
b.
c.
b.
c.
Charging
1.5 ohms and 1.0 ohms
5.5 ohms
1.09 A
0.364 A
Itoaster = 5.22 A,
Iiron = 10.44 A
Iprocessor = 4.3 A
Yes. The problem is that the fuse will blow because there will be too much current in the circuit.
22 ohms
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