Part Seven - Circuits Flashcards
What does an ammeter measure?
Flow of current through a component or circuit. (Amps)
What does a voltmeter measure?
The voltage drop over a component. (Volts)
What is the rule concerning current in a series circuit?
The same magnitude of current passes through each part of a series circuit
What is the rule concerning voltage in a series circuit?
The total voltage equals the sum of all voltage drops of all components in a series circuit.
What is the rule concerning resistance in a series circuit?
The resistance is equal to the sum of all resistances in a series circuit. R(total) = R1 + R2+ R3 + …
What is the rule concerning power in a series circuit?
The total power is the sum of the power of each resistor/component, in a series circuit.
What is the rule concerning resistance in a parallel circuit?
Total resistance is equal inverse sum of all resistances in a parallel circuit. 1/R(total) = 1/R1 + 1/R2 + 1/R3 + …
What is the rule concerning current in a parallel circuit?
The sum of the currents through each path is equal to the total current in the parallel circuit.
What is the rule concerning voltage in a parallel circuit?
The voltage is the same across each component of the parallel circuit.
What is the rule concerning power in a parallel circuit?
The total power is the sum of the power of each resistor/component in a parallel circuit.
In a series circuit with two resistors, question 55 on the review, R1=6Ω and R2=12Ω, and the voltage over R1 is 5 V, what is the current passing through R2?
I1 = V1/R1 I1 = 5V/6 ohms I1 = 0.83 A I1 = I2 = 0.83 A REMEMBER - in series circuits the magnitude of current through each resistor/component is equal.
In a series circuit with two resistors, question 55 on the review, R1=6Ω and R2=12Ω, and the voltage over R1 is 5 V, what is the voltage over R2?
V2 = (I2)(R2) V2 = (0.83 A)(12 ohms) V2 = 9.99 V
In a parallel circuit, two resistors in parallel, question 56, R1=5Ω and R2=4Ω, and the current passing through R1 is 0.1 A, what is the voltage over R2?
V1 = (R1)(I1) V1 = (5 ohms)(0.1 A) V1 = 0.5 V V1 = V2 = 0.5 V REMEMBER - for parallel circuits, the voltage drop across resistors in parallel are equal.
In a parallel circuit, two resistors in parallel, question 56, R1=5Ω and R2=4Ω, and the current passing through R1 is 0.1 A, what’s the current thru R2?
I2 = V2/R2 I2 = 0.5 V/4 ohms I2 = 0.125 A
What is the total/simplified resistance of the series circuit with a power source of 125 V and three resistors, R1 = 20 ohms, R2 = 30 ohms, R3 = 50 ohms, question 57?
R(total) = R1 + R2 + R3 R(total) = 20 ohms + 30 ohms + 50 ohms R(total) = 100 ohms
What is the total current of the series circuit with a power source of 125 V and three resistors, R1 = 20 ohms, R2 = 30 ohms, R3 = 50 ohms, question 57?
I(total) = V(total)/R(total) I(total) = 125 V/100 ohm I(total) = 1.25 A
What is the current through each resistor in the series circuit where R1=20Ω, R2=30Ω and R3=50Ω and the total volts =125, question 57?
Because the resistors are in series, so the magnitude of the current through each resistor is equal/constant.
I(total) = I1 = I2 = I3
I1 & I2 & I3 = 1.25 A
What is the voltage drop of each resistor in a series circuit in question 57, where R1=20Ω, R2=30Ω and R3=50Ω and the total volts =125?
V1 = (I1)(R1) V1 = (1.25 A)(20 ohms) V1 = 25 V
V2 = (I2)(R2) V2 = (1.25 A)(30 ohms) V2 = 37.5 V
V3 = (I3)(R3) V3 = (1.25 A)(50 ohms) V3 = 62.5 V
The voltage of each individual resistor added together should equal the voltage of the power supply - 27 V + 37.5 V + 62.5 V = 125 V
What is the power “used” by each resistor in the series circuit from problem 57, where R1=20Ω, R2=30Ω and R3=50Ω and the total volts =125?
P1 = (V1)(I1) P1 = (21 V)(1.25 A) P1 = 26.25 W
P2 = (V2)(I2) P2 = (37.5 V)(1.25 A) P2 = 46.88 W
P3 = (V3)(I3) P3 = (62.5 V)(1.25 A) P3 = 78.13 W
What is the total/simplified resistance of the parallel circuit from problem 58, where there are three parallel resistors, R1 = 20 ohms, R2 = 100 ohms, R3 = 50 ohms, and the power source has 125 V.
1/R(total) = 1/R1 + 1/R2 + 1/R3 1/R(total) = 1/20 ohms + 1/100 ohms + 1/50 ohms 1/R(total) = 5/100 ohms + 1/100 ohms + 2/100 ohms 1/R(total) = 8/100 ohms
NOW - you can solve for R(total) by multiplying R(total) to both sides and then dividing both sides by 8/100 - and to divide by a fraction you multiply the inverse.
So: 1 = 8/100 ohms (R(total))
(100/8)(1) = R(total)
R(total) = 12.5 ohms
OR - when you are at this stage: 1/R(total) = 8/100 ohms you know that the inverse is true, so you flip both and solve, getting the same result.
What is the total current of the parallel circuit from problem 58, where there are three parallel resistors, R1 = 20 ohms, R2 = 100 ohms, R3 = 50 ohms, and the power source has 125 V.
I1 = (V1)/(R1) I1 = (125 V)/(20 ohms) I1 = 6.25 A
I2 = (V2)/(R2) I2 = (125 V)/(100 ohms) I2 = 1.25 A
I3 = (V3)/(R3) I3 = (125 V)/(50 ohms) I3 = 2.50 A
The current of each individual resistor added together should equal the total current of the circuit - 6.25 A + 1.25 A + 2.50 A = 10 A
What is the total current of the parallel circuit from problem 58, where there are three parallel resistors, R1 = 20 ohms, R2 = 100 ohms, R3 = 50 ohms, and the power source has 125 V.
I(total) = (V1)/(R1) I(total) = (125 V)/(12.5 ohms) I(total) = 10 A
What is the voltage drop of each resistor in the parallel circuit from problem 58, where there are three parallel resistors, R1 = 20 ohms, R2 = 100 ohms, R3 = 50 ohms, and the power source has 125 V.
Because the resistors are in parallel, the voltage drops of each parallel resistor is equal in magnitude.
V(total) = V1 = V2 = V3 V1 = 125 V V2 = 125 V V3 = 125 V
What is the power “used” by each resistor in the parallel circuit from problem 58, where there are three parallel resistors, R1 = 20 ohms, R2 = 100 ohms, R3 = 50 ohms, and the power source has 125 V.
P1 = (V1)(I1) P1 = (125 V)(6.25 A) P1 = 781.25 W
P2 = (V2)(I2) P2 = (125 V)(1.25 A) P2 = 156.25 W
P3 = (V3)(I3) P3 = (125 V)(2.50 A) P3 = 312.50 W
