Question

3. a) A tungsten wire of length 20.0 m, initially at temperature 20.0°C, has a diameter of 2.00 mm. The potential difference across the whole wire is 180 V. Determine the current intensity in the wire b) Due to ohmic losses, the wire is heated up to 60.0°C, but the potential difference through the wire remains the same. Find the resulting current intensity. Note: po = 5,6x10 82 m, a=4.5x10(°C)

Answer 1

This problem can be solved by first calculating the resistance of the wire at 20 degree celsius using the resisitivity of tungsten and then finding the resistance of the wire at 60 degree celsius using the temperature co-efficient of resistance of the wire. The current intensity at 20 degree celsius came out as 504.91 Ampere and the current intensity at 60 degree celsius came out as 427.86 Amperes. Detailed solution given below. Ask me if you have any doubts. Feedback will be helpful.


A Solution (a) For a wire of length L, cross-section area A, and resistivity of the material used to make the wire to . Resistance of the wire Ŕ is given by: . R = pola In our problem, A= n d² where d is the diameter . of the wire. L= 20.om d=2.00 mm = 2.00 x 103 m [Imm= 10 m] Po = 5.6X108 -m As nga 3.1416 X (2.00 81034 m2 3.1416 x 100 m2 so, R.= PoL . 5.68108x202 A 8.1416 8 100 or R = 0.35651 . . Potential difference across the wire is V= 180 volt Using Ohm's law, v=iR , where is the current - intensity . ar iz = 504.91 Ampere cbs at temperature To= 20°c, we found the resistance to be Ro=0:35652 I .

Given that the temperature co-efficient of resistance is a the resistance R at a temperature T is given by R = Ro [+ x (T-To)] In our problem the wire heats up to 60°C from To=20°c. so, T=60°C Ro=0.35652 X= 4.5 X10-3 (°C) Se R = Ro [lta (T-T]] 120.3565 [1+ (4.5 x 10-3x(60-20)] = 0.3565 [1 +0.18]. 30.3565 % pour l'18 or R = 0.4207 s. Now, since the potential difference across the wire remains the same, V=180° Volt Ohm's law says that . New current intensity i= V = 180 WILL MANGLULUSLAUL R 0.4207 Lot U or i = 427. 86 Ampere Hence, the resultant current intensity is . 427,86 Ampere