How Do You Calculate RMS and Maximum EMF in Electromagnetic Induction Problems?

[AaronLosAl]

Homework Statement

I just got to two problems on my high school homework that i have no idea what to do because the teacher did not tell us how.

A generator can provide a maximum output emf of 170V. Calculate the rms potential difference.

Homework Equations

Do not know any relevant equations and i cannot find them in the book.

The Attempt at a Solution

170V(.707)=120V?


And then i do not even know where to start on this question.

Suppose you build a hydroelectric turbine in your back year along a river. The turbine is connected to a generator which can provide a rms potential difference of 320 V onto a 100 OHM load. Calculate the maximum output emf. Find the rms current and the maximum AC current in the curcuit.

Please help me!

 

[anathema]

Dear student,

I can understand your confusion and frustration with these problems. It is important for teachers to provide relevant equations and explanations in order for students to fully understand and complete their assignments. However, let me help you understand these concepts.

For the first problem, you are correct in using the equation V_rms = V_max(0.707) to find the rms potential difference. This equation is derived from the fact that the root mean square (rms) value of an AC voltage is 0.707 times the maximum value. Therefore, your answer of 120V is correct.

For the second problem, we can use Ohm's law to find the maximum output emf. Ohm's law states that V = IR, where V is the potential difference, I is the current, and R is the resistance. In this case, we know V = 320V and R = 100Ω. Therefore, the maximum output emf would be V = IR = (320V)(100Ω) = 32,000V.

To find the rms current, we can use the same equation V_rms = V_max(0.707). Rearranging this equation, we get I_rms = I_max(0.707). Since we know the maximum output emf is 32,000V, we can use Ohm's law again to find the maximum current. I_max = V/R = (32,000V)/(100Ω) = 320A. Therefore, the rms current would be I_rms = (320A)(0.707) = 226.24A.

Lastly, to find the maximum AC current, we need to know the peak-to-peak value of the current. This can be found by multiplying the rms current by 2√2. Therefore, the maximum AC current would be I_maxAC = (226.24A)(2√2) = 639.98A.

I hope this helps you understand these concepts better. Remember, it is always important to ask your teacher for clarification and additional resources if you are struggling with a concept. Keep up the good work!

 

[m2003]

Electromagnetic induction is the process of generating an electric current by moving a conductor through a magnetic field or by changing the magnetic field through a stationary conductor. This phenomenon is widely used in generators, which convert mechanical energy into electrical energy.

In order to solve the first problem, we need to use the equation for root mean square (rms) potential difference, which is given by V(rms) = V(max)/√2. So, in this case, V(rms) = 170V/√2 = 120V. Your attempt at the solution is correct.

For the second problem, we need to use the equation for maximum output emf, which is given by E(max) = V(rms) x √2. So, in this case, E(max) = 320V x √2 = 452.4V. This is the maximum potential difference that the generator can provide.

To find the rms current, we can use Ohm's law, which states that V = IR, where V is the potential difference, I is the current and R is the resistance. So, in this case, I(rms) = V(rms)/R = 320V/100Ω = 3.2A.

To find the maximum AC current, we need to use the equation I(max) = I(rms) x √2. So, in this case, I(max) = 3.2A x √2 = 4.5A. This is the maximum current that can flow through the circuit.

I hope this helps you understand electromagnetic induction better and solve your homework problems. Remember to always review the relevant equations and concepts in your textbook and seek help from your teacher if you are still unsure. Good luck!