Answer:
hello your question is incomplete attached below is the complete and the required circuit diagrams
answer :
Ai) <em>This proves that when the current across the resistor is doubled the value of the voltage across the resistor doubles as well</em>
B) It is better to take more measurements in the range 20mA < I < 40mA because of the amount of temperature reached by the bulb and the change in resistance is affected by the temperature
hence At 0 mA current, there won't be any noticeable change
Explanation:
Ai) The voltage across the resistor will double when you double the current through the resistor
Given that : V = I*R.
lets assume : I = 2 amperes , R = 3 ohms
V = 2*3 = 6 v
secondly lets assume double the value of (I) i.e. I = 4 amperes
hence : V = 4*3 = 12 volts
<em>This proves that when the current across the resistor is doubled the value of the voltage across the resistor doubles as well</em>
Aii) Showing the two data points from simulation
I1 V1 I2= 2I1 V2=2V1 V1/ I1 =V2/I2
0.9*10^3 9 * 10^3 1.8*10^3 18*10^3 10 ohms
1.6 * 10^3 16 * 10^3 3.2*10^3 32*10^3 10 ohms
B) It is better to take more measurements in the range 20mA < I < 40mA because of the amount of temperature reached by the bulb and the change in resistance is affected by the temperature
hence At 0 mA current, there won't be any noticeable change