A transformer's secondary coil contains 100 turns and operates at 10 volts. If the turns ratio is 18:1, find (a) the voltage ratio, (b) the primary voltage, and (c) the number of primary turns.

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5. A transformer's secondary coil contains 100 turns and operates at 10 volts. If the turns ratio is 18:1, find (a) the voltage ratio, (b) the primary voltage, and (c) the number of primary turns.

The turns ratio of a transformer is given by $\frac{N_p}{N_s}$, and for an ideal transformer, this is equal to the voltage ratio $\frac{V_p}{V_s}$.

a) Find the voltage ratio.

Step 1: State the given turns ratio. The turns ratio $\frac{N_p}{N_s}$ is $18:1$.

Step 2: Determine the voltage ratio. For an ideal transformer, the voltage ratio is equal to the turns ratio. $\frac{V_p}{V_s} = \frac{N_p}{N_s} = 18:1$ The voltage ratio is $\boxed{18:1}$.

b) Find the primary voltage.

Step 1: Use the voltage ratio and the secondary voltage. $\frac{V_p}{V_s} = 18$ Given $V_s = 10 \text{ V}$.

Step 2: Calculate the primary voltage. $V_p = 18 \times V_s = 18 \times 10 V = 180 \text{ V}$ The primary voltage is $\boxed{180 \text{ V}}$.

c) Find the number of primary turns.

Step 1: Use the turns ratio and the secondary turns. $\frac{N_p}{N_s} = 18$ Given $N_s = 100 \text{ turns}$.

Step 2: Calculate the number of primary turns. $N_p = 18 \times N_s = 18 \times 100 turns = 1800 \text{ turns}$ The number of primary turns is $\boxed{1800 \text{ turns}}$.

6. A 110-volt alternating current line is fed through a 55-turn step-down autotransformer. If you want a 28-volt output, find the secondary turns and the turn number to tap.

For an autotransformer, the ratio of voltages is equal to the ratio of the number of turns.

Step 1: Identify the given values. $V_p = 110 \text{ V}$ $N_p = 55 \text{ turns}$ $V_s = 28 \text{ V}$

Step 2: Calculate the number of secondary turns ($N_s$). $\frac{V_p}{V_s} = \frac{N_p}{N_s}$ $\frac{110 V}{28 V} = \frac{55 turns}{N_s}$ $N_s = \frac{28 V \times 55 turns}{110 V}$ $N_s = \frac{1540}{110} turns = 14 \text{ turns}$

Step 3: Determine the turn number to tap. For a step-down autotransformer, the secondary turns represent the tap point from the common end. The secondary turns are $\boxed{14 \text{ turns}}$ and the turn number to tap is $\boxed{14}$.

7. When two coils are put near together, a changing current in the first coil induces a voltage in the second. This phenomenon is called:

This phenomenon is called mutual induction.

8. Two inductive coils near each other have a mutual inductance of 0.6H. Current across one coil is boosted from 1A to 6A in 0.03 seconds. The voltage induced in the other coil is:

The magnitude of the induced voltage ($\mathcal{E}$) in the second coil due to a changing current in the first coil is given by $\mathcal{E} = M \frac{\Delta I}{\Delta t}$.

Step 1: Identify the given values and calculate the change in current. $M = 0.6 \text{ H}$ $\Delta I = 6 A - 1 A = 5 \text{ A}$ $\Delta t = 0.03 \text{ s}$

Step 2: Substitute the values into the formula. $\mathcal{E} = (0.6 H) \frac{5 A}{0.03 s}$

Step 3: Calculate the induced voltage. $\mathcal{E} = \frac{3}{0.03} V = 100 \text{ V}$ The voltage induced in the other coil is $\boxed{100 \text{ V}}$.

9. Two identical coils, A and B, with 400 turns set so that 60% of the flux produced by one coil connects to the other. Calculate the mutual inductance between coils A and B if a current of 10A runs through coil A, producing a flux of 20mWb.

Mutual inductance ($M$) is defined as $M = \frac{N_B \Phi_{BA}}{I_A}$, where $N_B$ is the number of turns in coil B, $\Phi_{BA}$ is the flux linked with coil B due to current $I_A$ in coil A.

Step 1: Identify the given values and calculate the flux linked with coil B. $N_A = N_B = 400 \text{ turns}$ $I_A = 10 \text{ A}$ $\Phi_A = 20 mWb = 20 \times 10^{-3} \text{ Wb}$ The flux linked with coil B is 60% of the flux produced by coil A: $\Phi_{BA} = 0.60 \times \Phi_A = 0.60 \times (20 \times 10^{-3} Wb) = 12 \times 10^{-3} \text{ Wb}$

Step 2: Substitute the values into the mutual inductance formula. $M = \frac{(400 turns) (12 \times 10^{-3} Wb)}{10 A}$

Step 3: Calculate the mutual inductance. $M = \frac{4.8 Wb}{10 A} = 0.48 \text{ H}$ The mutual inductance between coils A and B is $\boxed{0.48 \text{ H}}$.

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