Test 11.1 Dot Convention

Given the locations of the dots in the 2-coil system, it follows that

υ₂ = M (di₁/dt)
υ₂ = -M (di₁/dt)
υ₂ = (1/M) (di₁/dt)
υ₂ = -(1/M) (di₁/dt)

Test 11.2 Dot Convention

Given the locations of the dots in the 2-coil system, it follows that

υ₂ = M (di₁/dt)
υ₂ = -M (di₁/dt)
υ₂ = (1/M) (di₁/dt)
υ₂ = -(1/M) (di₁/dt)

Test 11.3 Equivalent Inductance

If L₁ = 10 mH, L₁ = 20 mH, and the mutual inductance is M = 4 mH, what is the equivalent inductance L_eq = ?

L_eq = 30 mH
L_eq = 26 mH
L_eq = 34 mH
L_eq = 22 mH

Test 11.4 T-Equivalent Circuit

Given that L₁ = L₂ = 20 mH and M = 2 mH,

L_x = L_y = L_z = 18 mH
L_x = L_y = 18 mH and L_z = 2 mH
L_x = L_y = 20 mH and L_z = 18 mH
L_x = L_y = L_z = 16 mH

Test 11.5 Ideal Transformer

The Thevenin-equivalent circuit consists of:

V_Th = 1.5e^j60° V and Z_Th = (0.5 - j2) Ω
V_Th = 1.5e^j60° V and Z_Th = (2 - j4) Ω
V_Th = -1.5e^j60° V and Z_Th = (0.5 - j1) Ω
V_Th = -1.5e^j60° V and Z_Th = (2 - j4) Ω

Test 11.6 Equivalent Inductance

The equivalent inductance at terminals (a,b) is:

L_eq = 19 H
L_eq = 23 H
L_eq = 21 H
L_eq = 17 H

Test 11.7 Ideal Transformer

Determine V_out.

V_out = 24∠0° V
V_out = 8∠0° V
V_out = 4∠0° V
V_out = 32∠0° V

Test 11.8 Power

How much power is delivered by the source? Voltages are rms.

P = 12 W
P = 18 mW
P = 24 W
P = 36 mW

Test 11.9 T-Equivalent Circuit

Given that L₁ = L₂ = 20 mH and M = 2 mH,

L_x = L_y = L_z = 18 mH
L_x = L_y = 18 mH and L_z = 2 mH
L_x = L_y = 22 mH and L_z = -2 mH
L_x = L_y = 22 mH and L_z = 2 mH

Test 11.10 Autotransformer

The autotransformer is used to step-down the voltage by a factor of 10. If N = 200, what should N₂ be?

N₂ = 90
N₂ = 10
N₂ = 20
N₂ = 80

Test 11.11 Autotransformer

The autotransformer is used to step-up the voltage by a factor of 20. If N = 200, what should N₂ be?

N₂ = 20
N₂ = 10
N₂ = 100
N₂ = 180

Test 11.12 Coupling Coefficient

What is the coupling coefficient of an ideal transformer?

k = 0
k = 0.5
k = 1
k = ∞

Test 11.13 MRI Superconducting Magnet

Why do Magnetic Resonance Imaging systems use superconducting magnets, as opposed to magnetic coils made of regular conductors?

Because superconducting magnets have very high resistance.
Because superconducting magnets have near-zero resistance.
Because superconducting magnets are lighter in weight.
Because superconducting magnets are cheaper.