Physics Lab 5 Practice Quiz

Four resistors 2.00 ohms, 5.00 ohms, 12.0 ohms and 15.0 ohms are placed in series with a 12.0 Volt battery. Determine the voltage drop across the 2.00 ohm resistor. 4.24 V 1.77 V 0.706 V 5.30 V

0.706 V Response Feedback: For a series circuit, the overall resistance (RTot) is simply the sum of the individual resistances. RTot = R1 + R2 + R3 + R4 RTot = 2.00 + 5.00 + 12.0 + 15.0 = 34.0 ohms The series of three resistors supplies a total or equivalent resistance of 34.0 Ohms. Since there is no branching, the current is the same through each resistor. This current is simply the overall current for the circuit and can be determined by finding the ratio of battery voltage to total resistance (VTot/RTot). ITot = VTot/RTot = (12.0 Volt) / (34.0 ohm) = 0.353 A The current through each of the resistors is 0.353 A. The voltage drop across each resistor is equal to the I * R product for each resistor. V1 = I1 * R1 = (0.353 A) * (2 ohm) = 0.706 V V2 = I2 * R2 = (0.353 A) * (5 ohm) = 1.77 V V3 = I3 * R3 = (0.353 A) * (12 ohm) = 4.24 V V4 = I4 * R4 = (0.353 A) * (15 ohm) = 5.30 V The voltage drop across the 2.00 ohm resistor = 0.706 V

Four resistors 2.00 ohms, 5.00 ohms, 12.0 ohms and 15.0 ohms are placed in parallel with a 12.0 Volt battery. Determine the current at 15.00 ohm resistor. 1.00 A 6.00 A 0.800 A 2.40 A

0.800 A Response Feedback: For a parallel circuit, the reciprocal of overall resistance (1 / RTot) is simply the sum of the reciprocals of individual resistances. That is 1 / RTot = 1 / R1 + 1 / R2 + 1 / R3 + 1 / R4 1 / RTot = 1 / 2 + 1 / 5 + 1 / 12 + 1 / 15 = 0.85 RTot = 1.18 ohms Since there is branching in this circuit, the total current will be equal to the sum of the currents at each resistor. The current at each resistor is the voltage drop across each resistor divided by the resistance of each resistor. The voltage drop across each resistor in this parallel circuit is same as the voltage supplied by the battery; 12 V. The currents through each resistor can be calculated as follows. I1 = V1 / R1 = (12.0 Volts) / (2.00 Ohms) = 6.00 A I 2 = V2 / R2 = (12.0 Volts) / (5.00 Ohms) = 2.40 A I 3 = V3 / R3 = (12.0 Volts) / (12.0 Ohms) = 1.00 A I 4 = V4 / R4 = (12.0 Volts) / (15.0 Ohms) = 0.800 A From above calculations, we can find that the current at 2.00 ohm resistor is 0.800 A

Four resistors 2.00 ohms, 5.00 ohms, 12.0 ohms and 15.0 ohms are placed in parallel with a 12.0 Volt battery. Determine the current at 12.00 ohm resistor. 1.00 A 2.40 A 6.00 A 0.800 A

1.00 A Response Feedback: For a parallel circuit, the reciprocal of overall resistance (1 / RTot) is simply the sum of the reciprocals of individual resistances. That is 1 / RTot = 1 / R1 + 1 / R2 + 1 / R3 + 1 / R4 1 / RTot = 1 / 2 + 1 / 5 + 1 / 12 + 1 / 15 = 0.85 RTot = 1.18 ohms Since there is branching in this circuit, the total current will be equal to the sum of the currents at each resistor. The current at each resistor is the voltage drop across each resistor divided by the resistance of each resistor. The voltage drop across each resistor in this parallel circuit is same as the voltage supplied by the battery; 12 V. The currents through each resistor can be calculated as follows. I1 = V1 / R1 = (12.0 Volts) / (2.00 Ohms) = 6.00 A I 2 = V2 / R2 = (12.0 Volts) / (5.00 Ohms) = 2.40 A I 3 = V3 / R3 = (12.0 Volts) / (12.0 Ohms) = 1.00 A I 4 = V4 / R4 = (12.0 Volts) / (15.0 Ohms) = 0.800 A From above calculations, we can find that the current at 2.00 ohm resistor is1.00 A

The positive terminals of the two voltage sources, V1 = 10 V and V2=20 V are connected in series with a 20 k ohm and 10 k ohm resistors. The negative terminals of both the voltage sources are interconnected through a 30 k ohm resistor. Determine the magnitude of the voltage at the 20 k ohm resistor. 10/3 V 5/3 V 5 V 30 V

10/3 V Response Feedback: According to KVL, the algebraic sum of the voltages around a closed loop equals zero. Assume that current in the circuit flows in the clockwise direction. Applying KVL, 20 k ohm * I +10 k ohm * I + 20V +30 k ohm * I - 10V = 0 10 V + 60000 *I = 0 I = -1/6 mA (negative sign indicates that current is flowing in counter clockwise direction) Voltage across 20 k ohm resistor = 20 k ohm * 1/6 mA = 10/3 V Voltage across 10 k ohm resistor = 10 k ohm * 1/6 mA = 5/3 V Voltage across 30 k ohm resistor = 30 k ohm * 1/6 mA = 5 V

Two resistors of values 10.0 ohms and 1000.0 ohms are connected in series with a voltage source of 12.0 V. What is the voltage drop across 1000.0 ohm resistor? 11.9 mA 12.0 mA 12.0 V 11.9 V 84.0 mA

11.9 V Response Feedback: V = IR; I = V/R = 12 V /(10 ohm + 1000 ohm) = 0.01188 A = 11.9 mA Since this is a series circuit, current is same at all points. 11.9 mA flows through 1000.0 ohm resistor. Therefore, Voltage dropped across it is given by V=IR = 11.9 mA* 1000.0 ohm = 11.9 V

Two resistors of values 10.0 ohms and 1000.0 ohms are connected in series with a voltage source of 12.0 V. What is the value of the current flowing through this circuit? 11.9 mA 12.0 mA 12.0 V 11.9 V 11.88 mA

11.9 mA Response Feedback: V = IR; I = V/R = 12 V /(10 ohm + 1000 ohm) = 0.01188 A = 11.9 mA (3 significant figures)

Two resistors of values 10.0 ohms and 1000.0 ohms are connected in series with a voltage source of 12.0 V. What is the value of the current flowing through 10.0 ohm resistor? 11.9 mA 12.0 mA 12.0 V 11.9 V 11.88 mA

11.9 mA Response Feedback: V = IR; I = V/R = 12 V /(10 ohm + 1000 ohm) = 0.01188 A = 11.9 mA (3 significant figures) Since the circuit is a series circuit with two resistors, same current passes through each of the resistors. Note that voltage drop across each resistor is different.

Two resistors of values 10.0 ohms and 1000.0 ohms are connected in series with a voltage source of 12.0 V. What is the voltage drop across 10.0 ohm resistor? 11.9 V 11.888 V 12.0 V 119 mV

119 mV Response Feedback: V = IR; I = V/R = 12 V /(10 ohms + 1000 ohms) = 0.01188 A = 11.9 mA Since this is a series circuit, current is same at all points. 11.9 mA flows through 10.0 ohm resistor. Therefore, Voltage dropped across it is given by V=IR = 11.9 mA* 10.0 ohms = 119 mV

A voltage source, a 12 k ohm and a 3 k ohm resistors are connected in parallel to each other. The current through the 3 k ohm resistor is measured to be 4mA. Determine the source voltage. 24 V 12 V 48 V 36 V

12 V

Four resistors 2.00 ohms, 5.00 ohms, 12.0 ohms and 15.0 ohms are placed in series with a 12.0 Volt battery. Determine the voltage drop across the 15.00 ohm resistor. 4.24 V 1.77 V 0.706 V 5.30 V

5.30 V Response Feedback: For a series circuit, the overall resistance (RTot) is simply the sum of the individual resistances. RTot = R1 + R2 + R3 + R4 RTot = 2.00 + 5.00 + 12.0 + 15.0 = 34.0 ohms The series of three resistors supplies a total or equivalent resistance of 34.0 Ohms. Since there is no branching, the current is the same through each resistor. This current is simply the overall current for the circuit and can be determined by finding the ratio of battery voltage to total resistance (VTot/RTot). ITot = VTot/RTot = (12.0 Volt) / (34.0 ohm) = 0.353 A The current through each of the resistors is 0.353 A. The voltage drop across each resistor is equal to the I * R product for each resistor. V1 = I1 * R1 = (0.353 A) * (2 ohm) = 0.706 V V2 = I2 * R2 = (0.353 A) * (5 ohm) = 1.77 V V3 = I3 * R3 = (0.353 A) * (12 ohm) = 4.24 V V4 = I4 * R4 = (0.353 A) * (15 ohm) = 5.30 V The voltage drop across the 2.00 ohm resistor = 0.706 V

In a loop in a closed circuit, the sum of the currents entering a junction equals the sum of the currents leaving a junction because The potential of the nearest battery is the potential at the junction There are no transformations of energy from one type to another in a circuit loop Capacitors tend to maintain current through them at a constant value Current is used up after it leaves a junction Charge is neither created nor destroyed at a junction

Charge is neither created nor destroyed at a junction

KCL is used when solving circuits with ... Closed Loops Sufficient Nodes / Junctions Capacitors None

Closed Loops

Two resistors R1 ohms and R2 ohms are connected in parallel with a voltage source of Vs. The current flowing through Vs is Is, R1 is I1 and R2 is I2. The potential drop across R1 is V1 and R2 is V2. Which of the following statements is true? Vs - V1 - V2 = 0 I1 - I2 = 0 Is - I1 - I2 = 0 Vs = (R1 + R2) (I1 + I2) Both a and c

Is - I1 - I2 = 0

Nodal Analysis applies the following principles KVL & Ohm's Law KCL & Ohm's Law KVL & Superposition KCL & Superposition

KCL & Ohm's Law

Which one of the following statements is most accurate? KVL is concerned with Voltage Drops KVL is calculated at Super Nodes only Both (A) and (B) None

KVL is concerned with Voltage Drops

Single loop circuits can be analyzed using _________ Kirchhoff's Voltage Law. Kirchhoff's Current Law. Both Kirchhoff's Voltage Law and Kirchhoff's Current Law. Either Kirchhoff's Voltage Law or Kirchhoff's Current Law. Ohm's law

Kirchhoff's Voltage Law

According to Kirchhoff's Second Law, in a given circuit the current is the same at all points because charge is conserved the current is the same at all points because charge is consumed the sum of the voltages about a closed path equals the sum of voltages supplied by the sources in that path the sum of the IR-drops across all resistances equals the internal resistance of the power supply the amount of current entering a junction is greater than the amount of current leaving the junction

the sum of the voltages about a closed path equals the sum of voltages supplied by the sources in that path

In general a series circuit divides ___ proportionally while a parallel circuit divides ___ proportionally. resistance..wattage resistance..voltage current..voltage voltage..current resistance..current

voltage..current Response Feedback: In general a series circuit divides VOLTAGE proportionally while a parallel circuit divides CURRENT proportionally.

Two resistors R1 ohms and R2 ohms are connected in series with a voltage source of Vs. The current flowing through Vs is Is, R1 is I1 and R2 is I2. The potential drop across R1 is V1 and R2 is V2. Which of the following statements is true? Vs - V1 - V2 = 0 I1 - I2 = 0 Is - I1 - I2 = 0 Vs = (R1 + R2) (I1 + I2) Both a and b

Both a and b Response Feedback: According to KVL, the algebraic sum of voltage drops across a circuit is zero. Applying KVL to this circuit, Vs - V1 - V2 = 0. Applying KCL at the node where R1 intersects with R2, we get current entering the node, I1 = current leaving the node, I2. Therefore, I1 - I2 = 0. Since this is a series circuit, Is = I1 = I2.

Two resistors R1 ohms and R2 ohms are connected in parallel with a voltage source of Vs. The current flowing through Vs is Is, R1 is I1 and R2 is I2. The potential drop across R1 is V1 and R2 is V2. Which of the following statements is true? Vs - V1 - V2 = 0 I1 - I2 = 0 Is - I1 + I2 = 0 V1 - V2 = 0

V1 - V2 = 0 Response Feedback: For a parallel circuit, voltage across each parallel path/ component is same. That is, for the given circuit, Vs = V1 = V2. According to KVL, the algebraic sum of voltage drops across a circuit is zero. Applying KVL to each loop of this circuit, Vs - V1 = 0, V1 - V2 = 0 and Vs - V2 = 0. To apply KCL at the node where the positive terminal of the voltage source intersects with R1 and R2, let us look at the directions of currents at that node. Current from the voltage source, Is enters the node and splits between R1 and R2. Therefore applying KCL, Is = I1 + I2 or Is - I1 - I2 = 0. Out of the multiple choice answers, only d is valid.

Two resistors R1 ohms and R2 ohms are connected in series with a voltage source of Vs. The current flowing through Vs is Is, R1 is I1 and R2 is I2. The potential drop across R1 is V1 and R2 is V2. According to Kirchhoff's Voltage Law, which of the following statements is true? Vs = (R1 + R2) (I1 + I2) V1 = I1 * R1 V2 = I2 * R2 Vs - V1 - V2 = 0 Is - I1 - I2 = 0

Vs - V1 - V2 = 0 Response Feedback: According to KVL, the algebraic sum of voltage drops across a circuit is zero. Applying KVL to this circuit, Vs - V1 - V2 = 0. Options b and c are correct statements according to Ohms Law; however, the question here is specifically asking for an equation that is true for the given circuit using KVL.

A ________ is defined as a place where two or more components are connected. Circuit Closed loop Node Voltage source Current source

Node

What is the significance of a negative sign from a calculation when solving circuit problems? None It means you did something wrong on your calculation Real resulting current or Voltage is in the opposite direction to one assumed You probably used a smaller scaling factor

Real resulting current or Voltage is in the opposite direction to one assumed

The algebraic sum of the changes of potential around any closed circuit path is Zero Maximum Zero if only there are no sources of emf in the path Maximum if there are no sources of emf in the path Equal to the sum of the currents in the branches of the path

Zero

According to Kirchhoff's First Law, when current divides at a junction the algebraic sum of the currents at the junction equals zero algebraic sum of the voltages about a closed path equals zero algebraic sum of the IR-drops across all circuit components equals the emf of the power supply algebraic sum of the currents at the junction equals the current delivered by the power supply sum of electric charge stored in each circuit component equals zero

algebraic sum of the currents at the junction equals zero

A ___________ in the network is any closed path through two or more elements of the network. loop Node Voltage source Current source

loop

According to Kirchhoff's First Law, in a given circuit the sum of the magnitudes of the currents entering any node is equal to the sum of the magnitudes of the currents leaving that node the current is the same at all points because charge is consumed the sum of the voltages about a closed path equals zero the sum of the IR-drops across all resistances equals the supply voltage the amount of current entering a junction is greater than the amount of current leaving the junction

the sum of the magnitudes of the currents entering any node is equal to the sum of the magnitudes of the currents leaving that node

According to Kirchhoff's Voltage/Second Law, the algebraic sum of the currents at the junction equals zero the voltages around a closed loop equals zero the IR-drops across all circuit components equals the emf of the power supply the currents at the junction equals the current delivered by the power supply electric charge stored in each circuit component equals zero

the voltages around a closed loop equals zero

Four 1.5 V AA batteries in series are used to power a transistor radio. If the batteries hold a total charge of 240.0 C, how long will they last if the radio has a resistance of 200.0 ohms? 2.2 hours 8.8 hours 130 hours 5.0 hours

2.2 hours Response Feedback: V=IR; I = Charge/time (C/s). Therefore, V = (Q/t)*R 4*1.5 V = (240 C / t) * 200 ohms; t = 8000 s = 8000/3600 = 2.22 hours

Four resistors 2.00 ohms, 5.00 ohms, 12.0 ohms and 15.0 ohms are placed in parallel with a 12.0 Volt battery. Determine the current at 5.00 ohm resistor. 1.00 A 2.40 A 6.00 A 0.800 A

2.40 A

What is the maximum number of 100 Watt lightbulbs you can connect in parallel in a 120 V home circuit without tripping the 20 A circuit breaker? 5 6 12 23 24

23 Response Feedback: V = 120 V, Maximum current flowing in the circuit = 20A. Power delivered by each light bulb = 100 W. P = V*I. Voltage dropped at each bulb, Vbulb = P/I = 100 W/ 20A = 5 V Voltage dropped at X number of bulbs = 120 V Number of bulbs that can be present in the circuit drawing a total of 20A = 120/5 = 24 Therefore, maximum number of bulbs that can be connected without tripping the circuit is 23.

Four resistors 2.00 ohms, 5.00 ohms, 12.0 ohms and 15.0 ohms are placed in series with a 12.0 Volt battery. Determine the current through the circuit. 1.58 A 0.353 V 2.83 A 0.353 A

0.353 A Response Feedback: For a series circuit, the overall resistance (RTot) is simply the sum of the individual resistances. RTot = R1 + R2 + R3 + R4 RTot = 2.00 + 5.00 + 12.0 + 15.0 = 34.0 ohms The series of three resistors supplies a total or equivalent resistance of 34.0 Ohms. Since there is no branching, the current is the same through each resistor. This current is simply the overall current for the circuit and can be determined by finding the ratio of battery voltage to total resistance (VTot/RTot). ITot = VTot/RTot = (12.0 Volt) / (34.0 ohm) = 0.353 A Therefore the current through the circuit and each of the resistors is same and is equal to 0.353 A.

A battery, a 12 k ohm and a 3 k ohm resistors are connected in parallel to each other. The current through the 3 k ohm resistor is measured to be 4mA. Determine the current though the 12 k ohm resistor. 1 mA 3 mA 4 mA 5 mA

1 mA Response Feedback: V3k = 3 k ohm * 4 mA = 12 V. Since this is a parallel circuit, voltage is same across each parallel component. Therefore, V3k = V12k = Vsource = 12 V. I12k = 12 V/12 k ohm = 1 mA

Four resistors 2.00 ohms, 5.00 ohms, 12.0 ohms and 15.0 ohms are placed in series with a 12.0 Volt battery. Determine the voltage drop across the 5.00 ohm resistor. 4.24 V 1.77 V 0.706 V 5.30 V

1.77 V Response Feedback: For a series circuit, the overall resistance (RTot) is simply the sum of the individual resistances. RTot = R1 + R2 + R3 + R4 RTot = 2.00 + 5.00 + 12.0 + 15.0 = 34.0 ohms The series of three resistors supplies a total or equivalent resistance of 34.0 Ohms. Since there is no branching, the current is the same through each resistor. This current is simply the overall current for the circuit and can be determined by finding the ratio of battery voltage to total resistance (VTot/RTot). ITot = VTot/RTot = (12.0 Volt) / (34.0 ohm) = 0.353 A The current through each of the resistors is 0.353 A. The voltage drop across each resistor is equal to the I * R product for each resistor. V1 = I1 * R1 = (0.353 A) * (2 ohm) = 0.706 V V2 = I2 * R2 = (0.353 A) * (5 ohm) = 1.77 V V3 = I3 * R3 = (0.353 A) * (12 ohm) = 4.24 V V4 = I4 * R4 = (0.353 A) * (15 ohm) = 5.30 V The voltage drop across the 2.00 ohm resistor = 0.706 V

Four resistors 2.00 ohms, 5.00 ohms, 12.0 ohms and 15.0 ohms are placed in parallel with a 12.0 Volt battery. Determine the current supplied by the battery. 1.00 A 10.2 A 6.00 A 0.800 A

10.2 A

Four resistors 2.00 ohms, 5.00 ohms, 12.0 ohms and 15.0 ohms are placed in series with a 12.0 Volt battery. Determine the voltage drop across the 12.00 ohm resistor. 4.24 V 1.77 V 0.706 V 5.30 V

4.24 V Response Feedback: For a series circuit, the overall resistance (RTot) is simply the sum of the individual resistances. RTot = R1 + R2 + R3 + R4 RTot = 2.00 + 5.00 + 12.0 + 15.0 = 34.0 ohms The series of three resistors supplies a total or equivalent resistance of 34.0 Ohms. Since there is no branching, the current is the same through each resistor. This current is simply the overall current for the circuit and can be determined by finding the ratio of battery voltage to total resistance (VTot/RTot). ITot = VTot/RTot = (12.0 Volt) / (34.0 ohm) = 0.353 A The current through each of the resistors is 0.353 A. The voltage drop across each resistor is equal to the I * R product for each resistor. V1 = I1 * R1 = (0.353 A) * (2 ohm) = 0.706 V V2 = I2 * R2 = (0.353 A) * (5 ohm) = 1.77 V V3 = I3 * R3 = (0.353 A) * (12 ohm) = 4.24 V V4 = I4 * R4 = (0.353 A) * (15 ohm) = 5.30 V The voltage drop across the 2.00 ohm resistor = 0.706 V

The positive terminals of the two voltage sources, V1 = 10 V and V2=20 V are connected in series with a 20 kilo ohms and 10 kilo ohms resistors. The negative terminals of both the voltage sources are interconnected through a 30 kilo ohms resistor. Determine the magnitude of the voltage at the 30 kilo ohms resistor. 10/3 V 5/3 V 5 V 30 V

5 V Response Feedback: According to KVL, the algebraic sum of the voltages around a closed loop equals zero. Assume that current in the circuit flows in the clockwise direction. Applying KVL, 20 k ohm * I+10 k ohm * I + 20V +30 k ohm * I - 10V = 0 10 V + 60000*I = 0 I = -1/6 mA (negative sign indicates that current is flowing in counter clockwise direction) Voltage across 20 k ohm resistor = 20 k ohm * 1/6 mA = 10/3 V Voltage across 10 k ohm resistor = 10 k ohm * 1/6 mA = 5/3 V Voltage across 30 k ohm resistor = 30 k ohm * 1/6 mA = 5 V

A battery, a 12 k ohm and a 3 k ohm resistors are connected in parallel to each other. The current through the 3 k ohm resistor is measured to be 4mA. Determine the source current. 1 mA 3 mA 4 mA 5 mA

5 mA Response Feedback: V3k = 3 k ohm * 4 mA = 12 V. Since this is a parallel circuit, voltage is same across each parallel component. Therefore, V3k = V12k = Vsource = 12 V. I3k = 4 mA; I12k = 12 V/12 k ohm = 1 mA I source = I3k + I12k = 4 mA + 1 mA = 5 mA

The positive terminals of the two voltage sources, V1 = 10 V and V2=20 V are connected in series with a 20 k ohm and 10 k ohm resistors. The negative terminals of both the voltage sources are interconnected through a 30 k ohm resistor. Determine the magnitude of the voltage at the 10 k ohm resistor. 5/3 V 5 V 30 V 10/3 V

5/3 V Response Feedback: According to KVL, the algebraic sum of the voltages around a closed loop equals zero. Assume that current in the circuit flows in the clockwise direction. Applying KVL, 20 k ohm * I + 10 k ohm * I+ 20V + 30 k ohm * I - 10V = 0 10 V + 60000 * I = 0 I = -1/6 mA (negative sign indicates that current is flowing in counter clockwise direction) Voltage across 20 k ohm resistor = 20 k ohm * 1/6 mA = 10/3 V Voltage across 10 k ohm resistor = 10 k ohm * 1/6 mA = 5/3 V Voltage across 30 k ohm resistor = 30 k ohm * 1/6 mA = 5 V

Four resistors 2.00 ohms, 5.00 ohms, 12.0 ohms and 15.0 ohms are placed in parallel with a 12.0 Volt battery. Determine the current at 2.00 ohm resistor. 0.800 A 2.40 A 1.00 A 6.00 A

6.00 A Response Feedback: For a parallel circuit, the reciprocal of overall resistance (1 / RTot) is simply the sum of the reciprocals of individual resistances. That is 1 / RTot = 1 / R1 + 1 / R2 + 1 / R3 + 1 / R4 1 / RTot = 1 / 2 + 1 / 5 + 1 / 12 + 1 / 15 = RTot = 1.18 ohms Since there is branching in this circuit, the total current will be equal to the sum of the currents at each resistor. The current at each resistor is the voltage drop across each resistor divided by the resistance of each resistor. The voltage drop across each resistor in this parallel circuit is same as the voltage supplied by the battery; 12 V. The currents through each resistor can be calculated as follows. I1 = V1 / R1 = (12.0 Volts) / (2.00 Ohms) = 6.00 A I 2 = V2 / R2 = (12.0 Volts) / (5.00 Ohms) = 2.40 A I 3 = V3 / R3 = (12.0 Volts) / (12.0 Ohms) = 1.00 A I 4 = V4 / R4 = (12.0 Volts) / (15.0 Ohms) = 0.800 A From above calculations, we can find that the current at 2.00 ohm resistor is 6.00 A