Regents Physics: Electric Circuits- SHS Community Service Quizlet

What is the current in a wire in which 600. C of change pass a point every 4.0 minutes?

Answer: 2.5 A

Explanation: The amount of charge flowing through the wire is given in coulombs (600. C) and the time is given in minutes (4.0). If the minutes are converted into seconds, the equation I= ∆q/t can be used to solve for I, the current. When the values are plugged in for the correct variables, I is equal to 2.5 A.

To make a wire with the lowest resistance, what material would you use? What material would you use for the highest resistance?

Answer: Lowest resistance- Silver, Highest resistance- Nichrome

Explanation: Material determines resistivity which determines resistance. The higher the resistivity, the higher the resistance. See the NYS Regents Physics Reference Table for a list of materials and their resistivities at 20 degrees Celsius.

A student measures a current of 0.10 ampere flowing through a lamp connected by short wires to 12.0 V source. What is the resistance of the lamp?

Answer: 120 Ω

Explanation: The givens in the question are the current (I= 0.10 A) and the voltage (V= 12.0 V), and we are being asked to solve for the resistance, R. This means we can use Ohm's Law as represented in the equation V=IR to solve for the resistance. When the correct values are plugged in for their designated variables and the equation is solved for R, a value of 120 Ω is obtained.

A light bulb is connected to a 1.5 volt cell that has a current of 28 mA. What is the power dissipated by the light bulb?

Answer: 4.2 x 10-² W

Explanation: The voltage (V= 1.5 V) and the current (I= 28 mA= 28 x 10-³ A) are our givens, and we are being asked to find the power of the lightbulb. This means we can use the equation P=IV to solve for the power, P. When the correct values are plugged in for their specified variables and you solve for P, 4.2 x 10-² W is the answer you obtain.

Three resistors, one with a resistance of 4.0 Ω, another with a resistance of 6.0 Ω, and the third with a resistance of 8.0 Ω are connected in series to a battery with a potential difference of 36 V. Find the equivalent resistance.

Answer: 18 Ω

Explanation: The resistances of the three resistors are given (4.0 Ω, 6.0 Ω, and 8.0 Ω). It is important to note that the resistors are connected in series, as that signifies that the equation we will be using to find the equivalent resistance is Req= R1+R2+R3... Using this equation, we can simply add the three resistances together to find the equivalent resistance, which is 18 Ω.

A 60. W light bulb and a 75. W light bulb operate from their own individual 120. V sources. Which bulb has a greater current running through it?

Answer: 75 W bulb

Explanation: The 75 W bulb will have a lower resistance and therefore more power if they are connected in parallel. If the light bulbs were in a series circuit, the bulbs would have the same amount of current running through it

When the resistance of a device is kept constant, how does the current running through it depend on the voltage?

Answer: There is a direct relationship between current and voltage, so the greater the voltage, the greater the current, and the smaller the voltage, the smaller the current

Explanation: Due to Ohm's Law (V= IR), it can be determined that there is a direct relationship between current and voltage. A direct relationship means that when one variable increases, the other variable increases as well and vice versa.

When the TV is turned on in the room of a house, another device is being added to the electric circuit that encompasses that room. Why do the other devices in the room not lose/gain any power?

Answer: Because they are connected in a parallel circuit

Explanation: In parallel circuits, the voltage of each device and the total voltage are equal, so the voltage will remain constant throughout a parallel circuit. Also, since the voltage and resistance of each individual device will remain the same, and current can be calculated by dividing voltage by resistance, the current for each device will remain the same as well. If the current and voltage are not changing, then the power for each device will not change either.

In an operating electrical circuit, the source of potential difference could be...

(1) voltmeter

(2) battery

(3) ammeter

(4) resistor

Answer: 2- a battery

Explanation: Potential difference is synonymous to voltage, which is the difference in electric potential energy between two points. This is what allows electric charge to move. This is provided by a battery or some other sort of device that can provide that difference in electric potential energy. A voltmeter simply measures voltage/potential difference, an ammeter measures current, and a resistor provides resistance for an electric charge, none of which actually provide a source of potential difference.

An aluminum wire of length 1.0 meter has a resistance of 9.0 × 10-3 ohm. If the wire were cut into two equal lengths, each length would have a resistance of

(1) 2.8 × 10-8 Ω

(2) 4.5 × 10-3 Ω

(3) 9.0 × 10-3 Ω

(4) 1.8 × 10-2 Ω

Answer: (2) 4.5 × 10-3 Ω

Explanation: One of the factors that affect the resistance of a conductor is length. The longer the wire, the more resistance it has. As a result, if a wire is cut in half, each half will only have half as much resistance as the original wire.

A lightbulb with a resistance of 2.9 ohms is operated using a 1.5-volt battery. At what rate is electrical energy transformed in the lightbulb? (1) 0.52 W

(2) 0.78 W

(3) 4.4 W

(4) 6.5 W

Answer: (2) 0.78 W

Explanation: The givens are the resistance (R= 2.9 Ω) and the voltage (V= 1.5 V). The question is at what rate is electrical energy transformed in the lightbulb, which is the definition of power. This means that we are calculating the power. Being that we are given the voltage and resistance, it makes the most sense to use the equation P=V²/R to solve for the power, P, which is 0.78 W.

A total charge of 100. coulombs flows past a fixed point in a circuit every 500. seconds. What is the current at this point in the circuit?

(1) 0.200 A

(2) 5.00 A

(3) 5.00 × 104 A

(4) 1.25 × 1018 A

Answer: (1) 0.200 A

Explanation: The givens are the total charge (∆q= 100. C) and the time in seconds (t= 500. s) and we are being asked for the current. This means that we should use the equation I= ∆q/t to solve for the current, I. When you divide 100. C/500. s, the answer is 0.200 A.

What is the maximum amount of work that a 6000.-watt motor can do in 10. seconds?

(1) 6.0 × 10¹ J

(2) 6.0 × 10² J

(3) 6.0 × 10³ J

(4) 6.0 × 10⁴ J

Answer: (4) 6.0 × 10⁴ J

Explanation: The givens are the power (6000 W) and the time (10 seconds) and we are being asked to find the work, W. This can be calculated using the equation W=Pt. When the power and time are multiplied together, a value of 6.0 × 10⁴ J is obtained for the work.

In a series circuit containing two lamps, the battery supplies a potential difference of 1.5 volts. If the current in the circuit is 0.10 ampere, at what rate does the circuit use energy?

(1) 0.015 W

(2) 0.15 W

(3) 1.5 W

(4) 15 W

Answer: (2) 0.15 W

Explanation: The question mentions that the lamps are in a series circuit, however, this is not relevant to the actual question. The question is asking for the rate at which the circuit uses energy, which refers to power. Being that our givens are the voltage (V= 1.5 V) and the current (I= 0.10 A), it makes the most sense to use the equation P= VI to calculate the power, P, which is 0.15 W.

A 6.0-ohm resistor that obeys Ohm's Law is connected to a source of variable potential difference. When the applied voltage is decreased from 12 V to 6.0 V, the current passing through the resistor

(1) remains the same

(2) is doubled

(3) is halved

(4) is quadrupled

Answer: (3) is halved

Explanation: The question is asking for the relationship between voltage and current. Due to Ohm's Law (V=IR), there is a direct relationship between voltage and current, meaning that if the voltage is halved in the problem, the current will be halved as well.

A 50-watt lightbulb and a 100-watt lighbulb are each operated at 110 volts. Compared to the resistance of the 50-watt bulb, the resistance of the 100-watt bulb is

(1) half as great

(2) twice as great

(3) one-fourth as great

(4) four times as great

Answer: (1) half as great

Explanation: The equation P= V²/R gives the relationship between power and resistance. This means that as the power increases, the resistance will decrease proportionally and vice versa, meaning that if the power is doubled, the resistance will be halved. You could also calculate the resistances for both lightbulbs and would have seen that the resistance for the 100-watt bulb was half as great

A device operating at a potential difference of 1.5 volts draws a current of 0.20 ampere. How much energy is used by the device in 60. seconds?

(1) 4.5 J

(2) 8.0 J

(3) 12 J

(4) 18 J

Answer: (4) 18 J

Explanation: The amount of electrical energy used, or work done, is being asked for in this question. Being that the voltage (1.5 V), current (0.20 A), and time (60. seconds) are given, it makes the most sense to use the equation W= VIt to solve for W, or electrical energy/work. When the correct values are plugged in for their designated variables, a value of 18 J is obtained for the energy, W.

As the number of resistors in a parallel circuit is increased, what happens to the equivalent resistance of the circuit and total current in the circuit?

(1) Both equivalent resistance and total current decrease.

(2) Both equivalent resistance and total current increase.

(3) Equivalent resistance decreases and total current increases. (4) Equivalent resistance increases and total current decreases.

Answer: (3) Equivalent resistance decreases and total current increases.

Explanation: The equation for finding the equivalent resistance in a parallel circuit is 1/Req= 1/R1+1/R2+1/R3... as more resistances are added, this number will become smaller and smaller. At the same time, the total current will increase because the even through the total voltage of the circuit will remain the same in a parallel circuit regardless of resistors, the total resistance is going to decrease because adding resistors is going to open up more paths for current to flow through and therefore make it easier for current to flow, causing the current to increase.

Two identical resistors connected in series have an equivalent resistance of 4 ohms. The same two resistors, when connected in parallel, have an equivalent resistance of

(1) 1 Ω

(2) 2 Ω

(3) 8 Ω

(4) 4 Ω

Answer: (1) 1 Ω

Explanation: The equation to find the equivalent resistance in series circuits is simply Req= R1+R2... Being that the question tells us that the resistors are identical, we can conclude that both resistors had resistances of 2 Ω. As a result, when we go to use the equivalent resistance equation for parallel circuits, 1/Req= 1/R1+1/R2, we are simply adding 1/2 +1/2 which is equal to 1 Ω

The current through a 10.-ohm resistor is 1.2 amperes. What is the potential difference across the resistor?

(1) 8.3 V

(2) 12 V

(3) 14 V

(4) 120 V

Answer: (2) 12 V

Explanation: The givens are the resistance, 10 ohms, and the current, 1.2 A. We are being asked to find the potential difference, which is synonymous to the voltage. We can use Ohm's Law, V=IR, to solve for the potential difference, V, by multiplying the resistance (10 ohms) and current (1.2 A) to obtain a value of 12 V.

A 4.50-volt personal stereo uses 1950 joules of electrical energy in one hour. What is the electrical resistance of the personal stereo?

(1) 433 Ω

(2) 96.3 Ω

(3) 37.4 Ω

(4) 0.623 Ω

Answer: (3) 37.4 Ω

Explanation: The givens are the voltage (4.5 V), the work/electrical energy (1950 J), and the time (1 hour), which needs to be converted into seconds (3600 seconds). We can use the equation W=V²t/R to solve for the resistance R, which comes out to 37.4 Ω

What is the total current in a circuit consisting of six operating 100-watt lamps connected in parallel to a 120-volt source?

(1) 5 A

(2) 20 A

(3) 600 A

(4) 12 000 A

Answer: (1) 5 A

Explanation: The power of each lamp (100 W) as well as the voltage (120 V) are given in the question, and we are being asked to find the total current in a parallel circuit. We can first find the current for one lamp using the equation P=VI, where we solve for I, which is 0.83. If we multiply that by 6 because there are six lamps and it is in a parallel circuit, so the total current is found by just adding all of the currents up, we obtain a total value of 5 A.

A circuit consists of a 10.0-ohm resistor, a 15.0-ohm resistor, and a 20.0-ohm resistor connected in parallel across a 9.00-volt battery. What is the equivalent resistance of this circuit?

(1) 0.200 Ω

(2) 1.95 Ω

(3) 4.62 Ω

(4) 45.0 Ω

Answer: (3) 4.62 Ω

Explanation: These resistors are connected in a parallel circuit, so in order to find the equivalent resistance we must use the correct formula for the equivalent resistance in a parallel circuit, which is 1/Req= 1/R1+1/R2+1/R3. When you plug into this equation correctly, a value of 4.62 Ω is obtained.

An electrical appliance draws 9.0 amperes of current when connected to a 120-volt source of potential difference. What is the total amount of power dissipated by this appliance?

(1) 13 W

(2) 110 W

(3) 130 W

(4) 1100 W

Answer: (4) 1100 W

Explanation: The givens are the current (9 A) and the voltage (120 V) and we are looking for the power, P. Based on what we are given, it makes the most sense to use the equation P= VI to find the power. When you multiply the voltage (120 V) by the current (9 A), a value of 1100 W is obtained.

An electric circuit contains a variable resistor connected to a source of constant voltage. As the resistance of the variable resistor is increased, the power dissipated in the circuit

(1) decreases

(2) increases

(3) remains the same

Answer: (1) decreases

Explanation: The equation P= V²/R demonstrates the relationship between power and voltage. When the voltage is kept constant, as the resistance increases, the power will decrease. You can also plug in arbitrary values for P, V, and R to understand this relationship with actual numbers.