Assignment A transformer has a primary coil of 200 turns and a secondary coil of 300 turns. If the voltage in the secondary is 100 V: a. Identify the type of transformer. b. Calculate the voltage on the primary coil. c. Find the current in the secondary coil if the primary coil has a current of 5 A.

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Answers

2 parts
b

66.67 V\text{66.67 V}

c

3.33 A\text{3.33 A}

Step-by-step solution

7 steps

Part a· 1 step

To identify the type of transformer, we compare the number of turns in the primary and secondary coils.

  1. Compare the number of turns.

    Given Np=200N_p = 200 turns and Ns=300N_s = 300 turns.
    Since Ns>NpN_s > N_p (300 turns > 200 turns), the transformer increases the voltage from the primary to the secondary.
    Therefore, it is a step-up transformer.

Part b

66.67 V\text{66.67 V}

To calculate the voltage on the primary coil, we use the transformer turns ratio formula.

  1. State the transformer voltage and turns ratio formula.

    VpVs=NpNs\frac{V_p}{V_s} = \frac{N_p}{N_s}

  2. Substitute the given values into the formula.

    We have Np=200N_p = 200 turns, Ns=300N_s = 300 turns, and Vs=100V_s = 100 V.

    Vp100 V=200300\frac{V_p}{100 \text{ V}} = \frac{200}{300}

  3. Solve for VpV_p.

    Vp=100 V×200300V_p = 100 \text{ V} \times \frac{200}{300} Vp=100 V×23V_p = 100 \text{ V} \times \frac{2}{3} Vp=2003 VV_p = \frac{200}{3} \text{ V} Vp≈66.67 VV_p \approx 66.67 \text{ V}

    The voltage on the primary coil is 66.67 V\boxed{\text{66.67 V}}.

Part c

3.33 A\text{3.33 A}

To find the current in the secondary coil, we assume an ideal transformer where the power in the primary coil equals the power in the secondary coil.

  1. State the power relationship for an ideal transformer.

    VpIp=VsIsV_p I_p = V_s I_s

  2. Substitute the known values into the equation.

    We have Vp=66.67V_p = 66.67 V (from part b), Ip=5I_p = 5 A, and Vs=100V_s = 100 V.

    (66.67 V)(5 A)=(100 V)Is(66.67 \text{ V})(5 \text{ A}) = (100 \text{ V}) I_s

  3. Solve for IsI_s.

    Is=(66.67 V)(5 A)100 VI_s = \frac{(66.67 \text{ V})(5 \text{ A})}{100 \text{ V}} Is=333.35 VA100 VI_s = \frac{333.35 \text{ VA}}{100 \text{ V}} Is=3.3335 AI_s = 3.3335 \text{ A}

    The current in the secondary coil is 3.33 A\boxed{\text{3.33 A}}.

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