Phase shift between voltage and current

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The discussion centers on determining the phase shift between voltage and current in a circuit with a 100Ω resistive load connected to a 100V power supply at 50Hz. It is established that if the load is purely resistive, there is no phase shift between voltage and current. Participants request the upload of the circuit diagram and the full problem statement for clarity. The original poster has attached the relevant equations but notes that the question did not include a circuit diagram. The conversation emphasizes the importance of having complete information to analyze phase relationships accurately.
IronaSona
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Homework Statement
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Relevant Equations
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Hello ,so am trying to find a phase shift between voltage and current ,I've done
Capture.PNG

what do i do after that ?and is this right ?

Q
A 100Ω load is connected to a power supply of 100V at 50Hz. At time, = 0, the instantaneous value for voltage across the load was zero and the current through the load was 0.7A.
 
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If the load is resistive, there is no phase shift. Can you upload the circuit that goes with this question? In fact, it would help if you could upload or write out the full problem statement. Thanks.

(use "Attach files" in the lower left to upload PDF or JPEG files)
 
berkeman said:
If the load is resistive, there is no phase shift. Can you upload the circuit that goes with this question? In fact, it would help if you could upload or write out the full problem statement. Thanks.

(use "Attach files" in the lower left to upload PDF or JPEG files)
done i have attached a full question, will all the equations which i have done so farm.
 

Attachments

berkeman said:
If the load is resistive, there is no phase shift. Can you upload the circuit that goes with this question? In fact, it would help if you could upload or write out the full problem statement. Thanks.

(use "Attach files" in the lower left to upload PDF or JPEG files)
and the question didnt come with a circuit
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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