a student connects the ends of a pencil lead to a 6.0V supply and measures the current to be 8.6A.the lead is rod of graphite of length 7.5 cm and diameter 1.4mm.
a)calculate the resistance of the pencil
b)use this data to calculate the resistivity of graphite
R = E/I = 6/8.6 = 0.70 ohms
Now use the definition of resistivity to work it out.
To calculate the resistance of the pencil lead, you can use Ohm's Law, which states that resistance (R) is equal to voltage (V) divided by current (I): R = V/I.
a) First, we need to determine the resistance. Given that the voltage (V) is 6.0V and the current (I) is 8.6A, we can substitute these values into the equation: R = 6.0V / 8.6A = 0.698Ω (rounded to three decimal places). Therefore, the resistance of the pencil lead is approximately 0.698Ω.
b) To calculate the resistivity of graphite, we can use the formula: ρ = RA/L, where ρ is the resistivity, R is the resistance, A is the cross-sectional area, and L is the length.
To find the cross-sectional area (A) of the pencil lead, we can use the formula for the area of a circle: A = πr^2, where r is the radius of the pencil lead. Given that the diameter is 1.4mm, we can calculate the radius (r) by dividing the diameter by 2: r = 1.4mm / 2 = 0.7mm = 0.07cm (since 1mm = 0.1cm).
Substituting the values into the formula for A, A = π(0.07cm)^2 = 0.15386 cm^2.
Lastly, we can substitute the values for R, A, and L (length = 7.5cm) into the resistivity formula: ρ = (0.698Ω)(0.15386cm^2) / 7.5cm = 0.014258 Ωcm (rounded to six decimal places). Therefore, the resistivity of graphite is approximately 0.014258 Ωcm.