How Six 2V lamps could be lit to their normal brightness.

AI Thread Summary
The discussion revolves around how to light six 2V lamps to their normal brightness, with the consensus that a parallel circuit is necessary for this configuration. The user expresses uncertainty about solving part B of the problem, suspecting it may involve Ohm's Law but struggling to apply it. They confirm their understanding of circuit drawing and potential difference calculations. The forum participants encourage the user to start a new thread for additional questions, adhering to forum rules. Overall, the conversation highlights the importance of circuit design and the application of electrical principles in achieving the desired lamp brightness.
Meezus
Messages
57
Reaction score
0

Homework Statement


[/B]
NgoAEq6.png

Homework Equations

The Attempt at a Solution



I'm not really sure how to attempt this question tbh. In order for the lamps to have normal brightness I think the circuit must be parallel. I also think maybe for A there can only be one lamp per branch because there's only 2V available, for B there can be 3 per row and for C it can just be series? [/B]
 
Physics news on Phys.org
Your attempt seems to be correct.
 
stockzahn said:
Your attempt seems to be correct.
Thank you :D I guess I did know haha. Any chance you could help with this?

3pZzg52.png

a) I know how to draw the circuit.
b) I'm completely stuck on B. I'm guessing it involves omhs law but I can't see where.
c) I can answer C with Potential difference (V) = current (I) x resistance (R).
d) I'm not sure how to work out this.
 
Meezus said:
Thank you :D I guess I did know haha. Any chance you could help with this?

3pZzg52.png

a) I know how to draw the circuit.
b) I'm completely stuck on B. I'm guessing it involves omhs law but I can't see where.
c) I can answer C with Potential difference (V) = current (I) x resistance (R).
d) I'm not sure how to work out this.
What is the p.d. across X?
 
haruspex said:
What is the p.d. across X?

Potential difference (V) = current (I) x resistance (R)

Potential difference (V) = 2 x Resistor X?
 
Meezus said:
Thank you :D I guess I did know haha. Any chance you could help with this?

3pZzg52.png

a) I know how to draw the circuit.
b) I'm completely stuck on B. I'm guessing it involves omhs law but I can't see where.
c) I can answer C with Potential difference (V) = current (I) x resistance (R).
d) I'm not sure how to work out this.

Yes, we'd like to help you with this, but at first please start a new thread with this problem. According to the forum rules only one problem should be discussed in each thread.
 

Similar threads

Back
Top