Saved Bookmarks
| 1. |
The temperature coefficient of resistivity for two materials A and B are 0.0031//""^@C , and 0.0068//""^@Crespectively. Two resistors R_1 and R_2, made from material A and B respectively, have resistances of 200 Omega and 100 Omega at 0^@C . Show the colour code of a carbon resistor that would have a resistance equal to the series combination of R_1 and R_2at a temperature of 100^@C(Neglect the ring corresponding to the tolerance of the carbon resistor). |
Answer» Solution : Here `(R_1)_0 = 200Omega , (R_2)_0 = 100 Omega , alpha_1 = 0.0031//""^@C , alpha_2 = 0.0068//""^@C " and" T = 100^@C` ` therefore (R_1)_T = (R_1)_0 [ 1+ alpha_1 (T - T_0) ]` ` = 200 [1 + 0.0031 xx 100] = 200 xx 1.31 = 262 Omega` and `(R_2)_T = (R_0)_0 [ 1+ alpha (T - T_0) ]` `= 100 [ 1+ 0.0068xx 100] = 100 xx 1.68 = 168 Omega` ` therefore ` Resistance of series combination at `100^@C` Hence as per colour CODE of carbon RESISTORS the code is as shown in Fig |
|