Numerical for JEE:

1. A copper wire has a resistance of 10 ohms at 20°C. What will be its resistance at 50°C? (Coefficient of resistivity of copper = 0.0043/°C)

Solution:

Using the formula: $$R_t = R_0 [1 + \alpha (t_f - t_i)]$$

Where,

• (R_t) = Resistance at temperature (t_f)
• (R_0) = Resistance at temperature (t_i)
• (\alpha) = Coefficient of resistivity
• (t_f) = Final temperature
• (t_i) = Initial temperature

Substituting values: $$R_{50} = 10 [1 + 0.0043 (50 - 20)]$$ $$R_{50} = 10 [1 + 0.0043 (30)]$$ $$R_{50} = 10 [1 + 0.129]$$ $$R_{50} = 10 (1.129)$$ $$R_{50} = 11.29 ohms$$

Therefore, the resistance of the copper wire at 50°C is 11.66 ohms.

2. A germanium semiconductor has a resistivity of 0.4 ohm-cm at room temperature (27°C). What will be its resistivity at 100°C? (Coefficient of resistivity of germanium = -0.0048/°C)

Solution:

Using the formula: $$\rho_t = \rho_0 [1 + \alpha (t_f - t_i)]$$

Where,

• (\rho_t) = Resistivity at temperature (t_f)
• (\rho_0) = Resistivity at temperature (t_i)
• (\alpha) = Coefficient of resistivity
• (t_f) = Final temperature
• (t_i) = Initial temperature

Substituting values: $$\rho_{100} = 0.4 [1 - 0.0048 (100 - 27)]$$ $$\rho_{100} = 0.4 [1 - 0.0048 (73)]$$ $$\rho_{100} = 0.4 [1 - 0.3504]$$ $$\rho_{100} = 0.4 (0.6496)$$ $$\rho_{100} = 0.25984 ohm-cm$$

Therefore, the resistivity of the germanium semiconductor at 100°C is 0.296 ohm-cm.

3. A nichrome wire has a resistance of 100 ohms at 25°C. What will be its resistance at 125°C? (Coefficient of resistivity of nichrome = 0.0004/°C)

Solution:

Using the formula: $$R_t = R_0 [1 + \alpha (t_f - t_i)]$$

Where,

• (R_t) = Resistance at temperature (t_f)
• (R_0) = Resistance at temperature (t_i)
• (\alpha) = Coefficient of resistivity
• (t_f) = Final temperature
• (t_i) = Initial temperature

Substituting values: $$R_{125} = 100 [1 + 0.0004 (125 - 25)]$$ $$R_{125} = 100 [1 + 0.0004 (100)]$$ $$R_{125} = 100 [1 + 0.04]$$ $$R_{125} = 100 (1.04)$$ $$R_{125} = 104 ohms$$

Therefore, the resistance of the nichrome wire at 125°C is 104.4 ohms.

Numerical for CBSE:

1. A carbon resistor has a resistance of 10 kΩ at 27°C. What will be its resistance at 77°C? (Coefficient of resistivity of carbon = -0.0005/°C)

Solution:

Using the formula: $$R_t = R_0 [1 + \alpha (t_f - t_i)]$$

Where,

• (R_t) = Resistance at temperature (t_f)
• (R_0) = Resistance at temperature (t_i)
• (\alpha) = Coefficient of resistivity
• (t_f) = Final temperature
• (t_i) = Initial temperature

Substituting values: $$R_{77} = 10 [1 - 0.0005 (77 - 27)]$$ $$R_{77} = 10 [1 - 0.0005 (50)]$$ $$R_{77} = 10 [1 - 0.025]$$ $$R_{77} = 10 (0.975)$$ $$R_{77} = 9.75 kΩ$$

Therefore, the resistance of the carbon resistor at 77°C is 9.23 kΩ.

2. A nichrome wire is used as the heating element in a toaster. If the wire has a resistance of 12 ohms at room temperature (25°C), what will be its resistance when the toaster is operating at 250°C? (Coefficient of resistivity of nichrome = 0.0004/°C)

Solution:

Using the formula: $$R_t = R_0 [1 + \alpha (t_f - t_i)]$$

Where,

• (R_t) = Resistance at temperature (t_f)
• (R_0) = Resistance at temperature (t_i)
• (\alpha) = Coefficient of resistivity
• (t_f) = Final temperature
• (t_i) = Initial temperature

Substituting values: $$R_{250} = 12 [1 + 0.0004 (250 - 25)]$$ $$R_{250} = 12 [1 + 0.0004 (225)]$$ $$R_{250} = 12 [1 + 0.09]$$ $$R_{250} = 12 (1.09)$$ $$R_{250} = 13.08 ohms$$

Therefore, the resistance of the nichrome wire when the toaster is operating at 250°C is 16 ohms.