Unit 19 - Electricity (All Math Problems)

Three resistances of 20 Ω, 50 Ω and 80 Ω make up a series circuit. The supply voltage is 400 V. The current flowing would be: A) 2.67 A B) 20 A C) 8 A D) 5 A E) 0.375 A

A) 2.67 A

The total equivalent resistance of three resistances in a series circuit can be given as: A) Rs = R₁ + R₂ + R₃ B) 1/Rp = (1/R₁) + (1/R₂) + (1/R₃) B) 1/Rp = (R₁/3) + (R₂/3) + (R₃/3) D) Rs = R₁ x R₂ x R₃ E) 1/Rp = (R₁ + R₂ + R₃) / 3

A) Rs = R₁ + R₂ + R₃

A transformer has a primary voltage of 500 volts and 20 turns on the primary winding. If the secondary winding has 5 turns, the voltage of the secondary is ________ volts. A) 2000 B) 2500 C) 125 D) 100 E) 4

C) 125

The equivalent resistance of three resistances of 10 Ω, 25 Ω and 50 Ω connected in series is _______ Ω. A) 0.16 B) 6.25 C) 85 D) 18.75 E) 8.5

C) 85

Find the applied voltage when three resistors of 40 Ω, 60 Ω, and 80 Ω are connected in parallel. The current passing through the 40 Ω resistance is 2.5 Amps. Also determine the line current and Wattage of the system.

1/Rp = (1/R₁) + (1/R₂) + (1/R₃) 1/Rp = 0.025 + 0.0167 + 0.0125 E = I*R E = 2.5 A * 40 Ω *E = 100 V* I₂ = 100/60 I₃ = 100/80 I = I₁ + I₂ + I₃ I = 2.5 + 1.666 + 1.25 *I = 5.416 A* *100 x 5.416 = 541.6 W*

Calculate the current flow in a circuit where the electromotive force is 6 Volts and the resistance is 6 Ohms.

I = E/R I = 1 Ampere

Three resistors are connected in parallel. R1 = 18 Ohms, R2 = 38 Ohms, R3 = 86 Ohms and the applied voltage is 150 Volts. The line current is: A) 2.6 Amps B) 4.5 Amps C) 8.6 Amps D) 10.6 Amps E) 14 Amps

D) 10.6 Amps

A 6 pole alternator producing 60 Hz power must turn at: A) 1800 rpm B) 2400 rpm C) 1000 rpm D) 1200 rpm E) 600 rpm

D) 1200 rpm

An electric motor develops 280 kW power output. The motor efficiency is 60%. If the supply voltage is 2.4 kV, determine the current flow of the motor in amps. A) 26.8 Amps B) 45.9 Amps C) 123.2 Amps D) 194.4 Amps E) 213.6 Amps

D) 194.4 Amps

An electric motor is connected to a 440 Volt line. The current drawn by the motor is 48 ampere. The power consumed by the motor is: A) 2.4 Watts B) 4.9 Watts C) 8.6 Watts D) 21.2 Watts E) 47.9 Watts

D) 21.2 Watts

What power is dissipated by a resistor with a supply voltage of 120 volts and current of 8 amperes? A) 960 Ω B) 15 Ω C) 15 watts D) 960 watts E) 96 kW

D) 960 watts * Volts x Amps = Watts

The power formula may be stated as: A) P = IR B) P = I/R C) P = E/R D) P = IE E) P = R/E

D) P = IE

Ohms Law can be stated as: A) E = IR B) R = I/E C) I = R/E D) volts equals ohms divided by amperes. E) watts times ohms equals volts.

A) E = IR ~or~ I = E/R ~or~ R = E/I E - electromotive force (volts) I - current (amperes) R - resistance (ohms)

Find the applied voltage when three resistors of 40 Ω, 60 Ω, and 80 Ω are connected in series. The current passing through the 40 Ω resistance is 2.5 Amps. Also determine the line current and Wattage of the system.

I = I₁ = I₂ = I₃ *I = 2.5 A* applied through each Rs = R₁ + R₂ + R₃ Rs = 40 + 60 + 80 Rs = 180 Ω Es = E₁ + E₂ + E₃ Es = (40x2.5) + (60x2.5) + (80x2.5) Es = 100 V + 150 V + 200 V *Es = 450 V* 450 / 180 = 2.5 A *450 x 2.5 = 1,125 W*

Find the voltage drop across a resistor with 10 Ω of resistance when a current of 3 Amps is flowing through it.

E = I*R E = 30 Volts

The voltage required to be supplied to a 12 kW motor drawing 24 A is: A) 5 kV B) 50 kV C) 288 V D) 2 kV E) 500 V

E) 500 Volts

The voltage required to force a current of 15 A through a resistance of 5 Ω will be _____ V. A) 0.33 V B) 3 V C) 7.5 V D) 33 V E) 75 V

E) 75 Volts

If Ep/Es = Np/Ns, then Ns will equal: A) Ns = Ep/NpEs B) Ns = EpEs/Np C) Ns = NpEsEp D) Ns = NpEp/Es E) Ns = NpEs/Ep

E) Ns = NpEs/Ep

When 1 Ampere of current flows through a resistance by the application of 1 Volt, the resistance will be 1 ___________. A) Watt B) Joule C) Kilowatt D) Ampere E) Ohm

E) Ohm

The basic unit of electrical power is the watt, which is the result of: A) volts times ohms. B) volts times amperes. C) ohms times amperes. D) ohms divided by amperes. E) volts divided by amperes.

B) volts times amperes. P = IE

The total equivalent resistance of three resistances in a parallel circuit can be given as: A) Rs = R₁ + R₂ + R₃ B) 1/Rp = (1/R₁) + (1/R₂) + (1/R₃) C) 1/Rp = (R1/1) + (R2/2) + (R3/3) D) Rs = R₁ x R₂ x R₃ E) 1/Rp = (R₁ + R₂ + R₃) / 3

B) 1/Rp = (1/R₁) + (1/R₂) + (1/R₃)

The equivalent resistance of three resistances of 10 Ω, 25 Ω and 50 Ω connected in parallel is _______ Ω. A) 0.16 B) 6.25 C) 85 D) 18.75 E) 8.5

B) 6.25