Math Problem Statement

Given:

VOC = 12 VRMS

Rint = 2.5 Ω

Rload = 30 Ω

a) Calculate Vload. Answer in V.

b) Calculate Iload. Answer in mA.

c) Calculate Pload. Answer in W.

d) Calculate the power required by the primary, Ppri. Answer in W.

e) Calculate the current into the primary Ipri. Answer in mA.

Solution

Given the following parameters:

  • VOC=12VRMSV_{OC} = 12 \, \text{V}_{RMS}
  • Rint=2.5ΩR_{int} = 2.5 \, \Omega
  • Rload=30ΩR_{load} = 30 \, \Omega

We can solve the questions step-by-step:

a) Calculate VloadV_{load} (in volts):

First, determine the total resistance in the circuit:

Rtotal=Rint+Rload=2.5Ω+30Ω=32.5ΩR_{total} = R_{int} + R_{load} = 2.5 \, \Omega + 30 \, \Omega = 32.5 \, \Omega

The load voltage VloadV_{load} can be found using the voltage divider formula:

Vload=VOC×RloadRtotal=12V×30Ω32.5ΩV_{load} = V_{OC} \times \frac{R_{load}}{R_{total}} = 12 \, \text{V} \times \frac{30 \, \Omega}{32.5 \, \Omega}

Vload11.08VV_{load} \approx 11.08 \, \text{V}

b) Calculate IloadI_{load} (in milliamps):

The load current IloadI_{load} can be found using Ohm's law:

Iload=VloadRload=11.08V30ΩI_{load} = \frac{V_{load}}{R_{load}} = \frac{11.08 \, \text{V}}{30 \, \Omega}

Iload0.369A=369mAI_{load} \approx 0.369 \, \text{A} = 369 \, \text{mA}

c) Calculate PloadP_{load} (in watts):

The power dissipated across the load can be calculated as:

Pload=Vload×Iload=11.08V×0.369AP_{load} = V_{load} \times I_{load} = 11.08 \, \text{V} \times 0.369 \, \text{A}

Pload4.09WP_{load} \approx 4.09 \, \text{W}

d) Calculate the power required by the primary, PpriP_{pri} (in watts):

The power required by the primary is the total power output, which includes the power dissipated by both RintR_{int} and RloadR_{load}.

First, calculate the total current:

Itotal=VOCRtotal=12V32.5ΩI_{total} = \frac{V_{OC}}{R_{total}} = \frac{12 \, \text{V}}{32.5 \, \Omega}

Itotal0.369AI_{total} \approx 0.369 \, \text{A}

The power required by the primary is:

Ppri=VOC×Itotal=12V×0.369AP_{pri} = V_{OC} \times I_{total} = 12 \, \text{V} \times 0.369 \, \text{A}

Ppri4.43WP_{pri} \approx 4.43 \, \text{W}

e) Calculate the current into the primary IpriI_{pri} (in milliamps):

We already calculated ItotalI_{total} in part (d), which is the same as IpriI_{pri}:

Ipri369mAI_{pri} \approx 369 \, \text{mA}


Do you want more details or have any questions?

Here are some related questions you might consider:

  1. What would happen if RloadR_{load} increased?
  2. How would the circuit be affected if RintR_{int} were negligible?
  3. What is the efficiency of the power transfer from the source to the load?
  4. How would the power dissipation change if the load resistance were halved?
  5. How could you reduce power losses in RintR_{int} in practical applications?

Tip: Always account for internal resistances in real circuits when estimating voltages and power efficiency.

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Math Problem Analysis

Mathematical Concepts

Electric Circuits
Ohm's Law
Voltage Divider Rule
Power in Electrical Circuits

Formulas

Ohm's Law: V = IR
Voltage Divider Rule: V_out = V_in * (R_load / R_total)
Power: P = VI

Theorems

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Suitable Grade Level

College Level