Question 1

In an ideal 1:8 step-down transformer, the primary power is 10 kW and the secondary current is 25 A.The primary voltage is:
A)25,600 V
B)3200 V
C)400 V
D)50 V
E)6.25 V

Accepted Answer

We know that power in the primary and secondary is equal. So, 
Primary power = Secondary power
10 kW = (25 A)^2 x R
R = (10,000 W) / (625 A^2) = 16 ohms

Also, we know that:
Vp / Vs = Np / Ns
Np / Ns = 1 / 8

So,
Vs / Vp = 8
Vs = 8 Vp

Now, using Ohm's Law in the secondary circuit:
Vs = Is x Rs
8 Vp = 25 A x 16 ohms
Vp = 3200 V

Question 2

A generator supplies 100 V to the primary coil of a transformer.The primary has 50 turns and the secondary has 500 turns.The secondary voltage is:
A)1000 V
B)500 V
C)250 V
D)100 V
E)10 V

Accepted Answer

The secondary voltage can be calculated using the transformer equation: $V_s = V_p \times \frac{N_s}{N_p}$, where $V_s$ is the secondary voltage, $V_p$ is the primary voltage, $N_s$ is the number of turns in the secondary coil, and $N_p$ is the number of turns in the primary coil. Substituting the given values: $V_s = 100 V \times \frac{500}{50} = 1000 V$.

Question 3

For a power transmission line, the transmission should be at low current and high voltage because:
A)this is the least dangerous to electrical workers
B)this gives the least dose of electromagnetic radiation to the public
C)this minimizes transmission losses
D)low current and high voltage is easier to transform than high current and low voltage
E)household appliances run at low current and high voltage

Accepted Answer

Transmission losses are proportional to current squared (I^2R), so minimizing current will minimize losses. High voltage is used to minimize losses due to impedance of the transmission line. The other options are not directly related to the choice of low current and high voltage transmission.

Question 4

A power transmission line carries 400A of current at a voltage of 765 kV.If the line has a resistance of 29 µΩ/m, what is the rate at which energy is being dissipated in 800 km of line?
A)0 W
B)3.7 kW
C)310 kW
D)3.7 MW
E)310 MW

Accepted Answer

The power dissipated in the line can be calculated using the formula P = I^2 * R, where I is the current and R is the total resistance. The total resistance R is the resistance per meter times the length of the line: R = 29 µΩ/m * 800,000 m = 23.2 Ω. Thus, P = (400 A)^2 * 23.2 Ω = 3.7 MW.

Question 5

The primary of an ideal transformer has 100 turns and the secondary has 600 turns.Then:
A)the power in the primary circuit is less than that in the secondary circuit
B)the currents in the two circuits are the same
C)the voltages in the two circuits are the same
D)the primary current is six times the secondary current
E)the frequency in the secondary circuit is six times that in the primary circuit

Accepted Answer

The primary current is six times the secondary current due to the transformer's turn ratio of 1:6 (100 turns to 600 turns), which inversely affects the current.

Question 6

A source with an impedance of 100 $\Omega$ is connected to the primary coil of a transformer and a resistance R is connected to the secondary coil.If the transformer has 500 turns in its primary coil and 100 turns in its secondary coil the greatest power will be dissipated in the resistor if R =
A)0 $\Omega$
B)0.25 $\Omega$
C)4.0 $\Omega$
D)20 $\Omega$
E)100$\Omega$

Accepted Answer

The impedance seen by the secondary coil of the transformer is given by Z_sec = Z_prim*(N_sec/N_prim)^2, where Z_prim is the impedance of the source, N_sec is the number of turns in the secondary coil, and N_prim is the number of turns in the primary coil. Substituting the given values, we get Z_sec = 100*(100/500)^2 = 1 F1F1F1S1F1F1F10.
The power dissipated in the resistor is given by P = V^2/R, where V is the voltage across the resistor. The voltage across the resistor is given by V = I*R, where I is the current in the circuit. The current in the circuit is given by I = V/Z_sec. Substituting the values, we get I = 10 F1F1F1V1F1F1F1S1F1F1F11 = 10 F1F1F1A1. Therefore, the power dissipated in the resistor is P = (10 F1F1F1V1F1F1F1S1F1F1F11)^2/R = 1000/R. For maximum power dissipation in the resistor, we need to minimize R. Therefore, the answer is the smallest value of R given, which is 4.0 F1F1F1S1F1F1F10.

Question 7

A step-down transformer is used to:
A)increase the power
B)decrease the power
C)increase the voltage
D)decrease the voltage
E)change ac to dc

Accepted Answer

Step-down transformers are designed to reduce the voltage of an alternating current (AC) power supply. This is done by having a greater number of turns in the primary coil of wire than the secondary coil. As current flows through the primary coil, it creates a magnetic field, which induces an electrical current in the secondary coil. The voltage of the induced electrical current will be lower than that of the original voltage in the primary coil.

Question 8

The resistance of the primary coil of a well-designed, 1:10 step-down transformer is 1 $\Omega$ .With the secondary circuit open, the primary is connected to a 12 V ac generator.The primary current is:
A)essentially zero
B)about 12 A
C)about 120 A
D)depends on the actual number of turns in the primary coil
E)depends on the core material

Accepted Answer

With the secondary circuit open, there is no load on the transformer, which means no current flows in the secondary coil. Hence, according to the transformer law, the primary current is also zero. The resistance of the primary coil does not affect the current in this case.

Question 9

The primary of a 3:1 step-up transformer is connected to a source and the secondary is connected to a resistor R.The power dissipated by R in this situation is P.If R is connected directly to the source it will dissipate a power of:
A)P/9
B)P/3
C)P
D)3P
E)9P

Accepted Answer

In a 3:1 step-up transformer, the voltage across the secondary is three times the primary voltage. If R is connected directly to the source, the voltage across it will be one-third of what it was when connected to the secondary. Since power is proportional to the square of the voltage (P = V^2/R), the power dissipated will be (1/3)^2 = 1/9 of the original power P.

Quiz 31: Electromagnetic Oscillations and Alternating Current

In an ideal 1:8 step-down transformer, the primary power is 10 kW and the secondary current is 25 A.The primary voltage is:

A generator supplies 100 V to the primary coil of a transformer.The primary has 50 turns and the secondary has 500 turns.The secondary voltage is:

For a power transmission line, the transmission should be at low current and high voltage because:

A power transmission line carries 400A of current at a voltage of 765 kV.If the line has a resistance of 29 µΩ/m, what is the rate at which energy is being dissipated in 800 km of line?

The primary of an ideal transformer has 100 turns and the secondary has 600 turns.Then:

A source with an impedance of 100 Ω\OmegaΩ is connected to the primary coil of a transformer and a resistance R is connected to the secondary coil.If the transformer has 500 turns in its primary coil and 100 turns in its secondary coil the greatest power will be dissipated in the resistor if R =

A step-down transformer is used to:

The resistance of the primary coil of a well-designed, 1:10 step-down transformer is 1 Ω\OmegaΩ .With the secondary circuit open, the primary is connected to a 12 V ac generator.The primary current is:

The primary of a 3:1 step-up transformer is connected to a source and the secondary is connected to a resistor R.The power dissipated by R in this situation is P.If R is connected directly to the source it will dissipate a power of:

A source with an impedance of 100 $\Omega$ is connected to the primary coil of a transformer and a resistance R is connected to the secondary coil.If the transformer has 500 turns in its primary coil and 100 turns in its secondary coil the greatest power will be dissipated in the resistor if R =

The resistance of the primary coil of a well-designed, 1:10 step-down transformer is 1 $\Omega$ .With the secondary circuit open, the primary is connected to a 12 V ac generator.The primary current is: