advanced in phys.

[رؤى عبدالله]

A power station is connected to a factory 2.5 km away by two wires. the potential difference between the terminals of the wires at the station is 240v and that for the other terminals at the factory is 220v. = calculate the resistance of one meter of the wire and its radius ifits resitivity 1.57x10-3ohm.m and the current is 80A

you have 6 resistores of resistamces1,2,2,4,5,6 .show with drawing how to connect them to obtain atotal resistance = 1ohm

A12 v battery has an internal resistance of 0.5ohm. calculate the percent of voltage drop for this battery when connected to a 2 ohm lamp

Abattery connected with a resistor(R) so the current intensity is (I) when another resistance (0.5R) connected with the first the current inc. to double . calculate the r in terms of R

[hatemT]

المشاركة الأصلية كتبت بواسطة رؤى عبدالله

A power station is connected to a factory 2.5 km away by two wires. the potential difference between the terminals of the wires at the station is 240v and that for the other terminals at the factory is 220v. = calculate the resistance of one meter of the wire and its radius ifits resitivity 1.57x10-3ohm.m and the current is 80A

The solution:
as we know; the resistance is equal to=V/I=(240-220)/80=0.25 ohm; thus the resistance per meter= 0.25/2500 ohm/m. "don't forget to multiply the ans. by2"
and then we can get the raduis fron the equation of the resistivity; R=(rho*L)/A, and A=Pi.r^2


you have 6 resistores of resistamces1,2,2,4,5,6 .show with drawing how to connect them to obtain atotal resistance = 1ohm
The solution:

Is lefted to yours

A12 v battery has an internal resistance of 0.5ohm. calculate the percent of voltage drop for this battery when connected to a 2 ohm lamp
The solution:
firstly before connecting with the resis. the voltage drop was equal to 12v" open circuit", and after connecting it; thus the current would be produced; as a resulte the voltage across the battery will fall to 12-(0.5.I); and as the current could be calcultad easily, from I=12/2.5=4.8 Amp. , soooooo this's
the voltage drop becomes 9.6v. ; the % voltage drop=(12-9.6)/12=20%

Abattery connected with a resistor(R) so the current intensity is (I) when another resistance (0.5R) connected with the first the current inc. to double . calculate the r in terms of R
The solution:
I think this's easy, you can answer it

Good luck; bye bye