1.
Suppose you double the voltage in a simple dc circuit, and cut the resistance in half. The current will become:
Correct Answer
A. Four times as great
Explanation
When the voltage in a simple DC circuit is doubled, and the resistance is halved, according to Ohm's Law (V=IR), the current in the circuit will increase. This is because the voltage directly affects the current, and when the voltage is doubled, the current will also double. Additionally, when the resistance is halved, the current will increase even more. Therefore, the current will become four times as great as it was before.
2.
A wiring diagram would most likely be found in:
Correct Answer
C. The service/repair manual for a radio receiver
Explanation
A wiring diagram is a visual representation of the electrical connections and components in a system. It provides detailed information about how the system is wired and helps in troubleshooting and repairing. Therefore, it is most likely to be found in the service/repair manual for a radio receiver, as this manual would contain specific instructions and diagrams for repairing the electrical components of the radio receiver.
3.
Given a dc voltage source delivering 24 V and a circuit resistance of 3.3 KΩ, what is the current?
Correct Answer
D. 7.3 mA
Explanation
The current can be calculated using Ohm's Law, which states that current (I) is equal to voltage (V) divided by resistance (R). In this case, the voltage is 24 V and the resistance is 3.3 KΩ (or 3300 Ω). By substituting these values into the formula, we get I = 24 V / 3300 Ω = 0.0073 A, which is equivalent to 7.3 mA.
4.
Suppose that a circuit has 472 Ω of resistance and the current is 875 mA. Then the source voltage is:
Correct Answer
A. 413 V
Explanation
The source voltage can be calculated using Ohm's Law, which states that voltage (V) is equal to current (I) multiplied by resistance (R). In this case, the current is given as 875 mA, which is equivalent to 0.875 A, and the resistance is given as 472 Ω. Therefore, the source voltage can be calculated as V = I * R = 0.875 A * 472 Ω = 411.5 V. Rounding this to the nearest whole number gives us the answer of 413 V.
5.
The dc voltage in a circuit is 550 mV and the current is 7.2 mA. Then the resistance is:
Correct Answer
B. 76 Ω
Explanation
The resistance can be calculated using Ohm's law, which states that resistance is equal to the voltage divided by the current. In this case, the voltage is given as 550 mV (or 0.55 V) and the current is given as 7.2 mA (or 0.0072 A). Dividing the voltage by the current gives us a resistance of approximately 76 Ω. Therefore, the correct answer is 76 Ω.
6.
Given a dc voltage source of 3.5 kV and a circuit resistance of 220 Ω, what is the current?
Correct Answer
D. None of the above
7.
A source delivers 12 V and the current is 777 mA. Then the best expression for the resistance is:
Correct Answer
A. 15 Ω
Explanation
The best expression for resistance can be determined using Ohm's Law, which states that resistance (R) is equal to the voltage (V) divided by the current (I). In this case, the voltage is 12 V and the current is 777 mA (or 0.777 A). Dividing 12 V by 0.777 A gives a result of approximately 15.4 Ω. However, since the answer choices provided do not include this exact value, the closest option is 15 Ω.
8.
A circuit has a total resistance of 473,332 Ω and draws 4.4 mA. The best expression for the voltage of the source is:
Correct Answer
C. 2.1 kV
Explanation
The best expression for the voltage of the source is 2.1 kV. This can be determined using Ohm's Law, which states that voltage (V) is equal to current (I) multiplied by resistance (R). Rearranging the formula, V = I * R, we can substitute the given values of 4.4 mA (0.0044 A) for current and 473,332 Ω for resistance. Multiplying these values together, we get V = 0.0044 A * 473,332 Ω = 2082.656 V, which can be rounded to 2.1 kV.
9.
The voltage is 250 V and the current is 8.0 mA. The power dissipated by the potentiometer is:
Correct Answer
C. 2.0 W
Explanation
The power dissipated by a device can be calculated using the formula P = IV, where P is power, I is current, and V is voltage. In this case, the current is given as 8.0 mA, which is equivalent to 0.008 A, and the voltage is given as 250 V. By substituting these values into the formula, we can calculate the power dissipated by the potentiometer as 2.0 W.
10.
The voltage from the source is 12 V and the potentiometer is set for 470 Ω. The power is about:
Correct Answer
A. 310 mW
Explanation
The power can be calculated using the formula P = V^2/R, where V is the voltage and R is the resistance. In this case, the voltage is 12 V and the resistance is 470 Ω. Plugging in these values, we get P = (12^2) / 470 = 144 / 470 ≈ 0.3064 W. Converting this to milliwatts, we get 0.3064 W * 1000 = 306.4 mW. Therefore, the power is approximately 310 mW.
11.
The current through the potentiometer is 17 mA and its value is 1.22KΩ. The power is:
Correct Answer
D. 350 mW
Explanation
The power can be calculated using the formula P = I^2 * R, where P is the power, I is the current, and R is the resistance. In this case, the current is given as 17 mA (which is 0.017 A) and the resistance is 1.22KΩ (which is 1220 Ω). Plugging these values into the formula, we get P = (0.017 A)^2 * 1220 Ω = 0.003674 W. Converting this to milliwatts, we get 0.003674 W * 1000 = 3.674 mW. Therefore, the correct answer is 350 mW.
12.
Suppose six resistors are hooked up in series, and each of them has a value of 540 Ω. Then the total resistance is:
Correct Answer
B. 3.24 KΩ
Explanation
When resistors are connected in series, their resistances add up to give the total resistance. In this case, there are six resistors, each with a value of 540 Ω. Therefore, the total resistance is 6 * 540 Ω = 3240 Ω = 3.24 KΩ.
13.
Four resistors are connected in series, each with a value of 4.0 KΩ. The total resistance is:
Correct Answer
D. 16 KΩ
Explanation
When resistors are connected in series, their resistances add up to give the total resistance. In this case, since all four resistors have a value of 4.0 KΩ, the total resistance is 4.0 KΩ + 4.0 KΩ + 4.0 KΩ + 4.0 KΩ = 16.0 KΩ.
14.
Suppose you have three resistors in parallel, each with a value of 68,000 Ω. Then the total resistance is:
Correct Answer
B. 23 KΩ
Explanation
When resistors are connected in parallel, the total resistance is calculated using the formula: 1/R_total = 1/R1 + 1/R2 + 1/R3 + ... In this case, we have three resistors with a value of 68,000 Ω each. Substituting the values into the formula, we get 1/R_total = 1/68,000 + 1/68,000 + 1/68,000. Simplifying this equation, we find that 1/R_total = 3/68,000. Taking the reciprocal of both sides, we get R_total = 68,000/3, which is approximately equal to 22,666.67 Ω. When rounded to the nearest kilo-ohm, the total resistance is 23 KΩ.
15.
There are three resistors in parallel, with values of 22 Ω, 27Ω, and 33 Ω. A 12-V battery is connected across this combination, as shown in Fig. 4-11. What is the current drawn from the battery by this resistance combination?
Correct Answer
A. 1.3 A
Explanation
When resistors are connected in parallel, the total resistance is given by the formula 1/Rt = 1/R1 + 1/R2 + 1/R3. Plugging in the values of the resistors (22, 27, and 33), we can calculate the total resistance. Using Ohm's Law, I = V/R, where V is the voltage (12V) and R is the total resistance, we can calculate the current drawn from the battery. In this case, the current is calculated to be 1.3 A.
16.
Three resistors, with values of 47 Ω, 68 Ω, and 82 Ω, are connected in series with a 50-V dc generator, as shown in Fig. 4-12. The total power consumed by this network of resistors is:
Correct Answer
C. 13 W
Explanation
The total power consumed by a network of resistors connected in series can be calculated using the formula P = V^2 / R, where P is the power, V is the voltage, and R is the total resistance. In this case, the total resistance is the sum of the resistances of the three resistors, which is 47 + 68 + 82 = 197 Ω. The voltage is given as 50 V. Substituting these values into the formula, we get P = (50^2) / 197 ≈ 12.7 W. Rounded to the nearest whole number, the total power consumed by this network of resistors is 13 W.
17.
You have an unlimited supply of 1-W, 100-Ω resistors. You need to get a 100- Ω, 10-W resistor. This can be done most cheaply by means of a series-parallel matrix of
Correct Answer
C. 4 X 4 resistors
Explanation
To achieve a 100-Ω, 10-W resistor, you need to combine resistors in a way that their resistance adds up to 100-Ω and their power rating adds up to 10-W. The 4 X 4 resistors configuration would be the most suitable option because it allows for more combinations of resistors in series and parallel. With 4 X 4 resistors, you can create various combinations of series and parallel connections to achieve the desired resistance and power rating, making it the most cost-effective choice.
18.
You have an unlimited supply of 1-W, 1000-Ω resistors, and you need a 500-Ω resistance rated at 7 W or more. This can be done by assembling
Correct Answer
A. Four sets of two 1000-Ω resistors in series, and connecting these four sets in parallel
Explanation
By connecting four sets of two 1000-Ω resistors in series, each set will have a total resistance of 2000-Ω. When these four sets are connected in parallel, the total resistance will be 500-Ω (2000-Ω/4). The power rating of each resistor is 1-W, so when four sets are connected in parallel, the total power rating will be 4-W (1-W x 4). This configuration meets the requirement of a 500-Ω resistance rated at 7 W or more.
19.
You have an unlimited supply of 1-W, 1000-Ω resistors, and you need to get a 3000-Ω, 5-W resistance. The best way is to:
Correct Answer
D. Do something other than any of the above
20.
Good engineering practice usually requires that a series-parallel resistive network be made:
Correct Answer
B. From resistors that are all the same
Explanation
A series-parallel resistive network is typically made from resistors that are all the same because this ensures that the resistors have the same resistance value. This allows for easier calculations and analysis of the network. Additionally, using resistors that are all the same ensures that the current is evenly distributed among the resistors, preventing any imbalances or excessive current flow through a single resistor. This helps to maintain the stability and reliability of the network.