A simple circuit is made of a power source, a rheostat and an ohmic resistor. The potential difference across the resistor is initially 5.0 ± 0.1 V. A good lab student measures the current across the resistor to be 1.65 ± 5 mA.

a)If the student increases the voltage across the resistor to 9.0 ± 0.1 V, around what value should the measurement of the current across the resistor be (include uncertainties)?

To find the current across the resistor when the voltage is increased to 9.0 ± 0.1 V, we can use Ohm's law. Ohm's law states that the current (I) flowing through a conductor is directly proportional to the voltage (V) applied across it and inversely proportional to the resistance (R) of the conductor.

Mathematically, Ohm's law can be expressed as:

I = V / R

where I is the current, V is the voltage, and R is the resistance.

In this scenario, the voltage across the resistor is 5.0 ± 0.1 V, and the current across the resistor is 1.65 ± 5 mA.

First, let's convert the current to amperes by dividing it by 1000:
1.65 mA = 1.65 / 1000 A = 0.00165 A

Next, let's calculate the resistance using Ohm's law:

R = V / I

Substituting the given values:
R = (5.0 ± 0.1 V) / (0.00165 ± 0.005 A)

Now, let's calculate the resistance using the given values:
R = (5.0 ± 0.1) / (0.00165 ± 0.005) = 3030.30 ± 759.49 ohms

So, the resistance of the resistor is approximately 3030.30 ± 759.49 ohms.

Now, we can use the resistance to find the current when the voltage is increased to 9.0 ± 0.1 V:

I = V / R

Substituting the values:
I = (9.0 ± 0.1 V) / (3030.30 ± 759.49 ohms)

Calculating the current using the given values:
I = (9.0 ± 0.1) / (3030.30 ± 759.49) = 0.00297 ± 0.00057 A

Converting the current back to milliamperes:
0.00297 A = 0.00297 * 1000 mA = 2.97 ± 0.57 mA

Therefore, when the voltage is increased to 9.0 ± 0.1 V, the current across the resistor should be approximately 2.97 ± 0.57 mA.