Final answer:
To find the current in the resistor, subtract the lower voltage from the upper voltage to find the voltage drop and then use Ohm's Law. For the given values, the current is approximately 14.89 mA. The accuracy of the current is ±2.02 mA, considering the uncertainties of voltage and resistance measurements.
Step-by-step explanation:
To calculate the level of current in the resistor with its accuracy, we first find the voltage drop across the resistor. The voltage drop (V) is the difference between the two voltage measurements: V = V1 - V2 = 12V - 5V = 7V. Now, Ohm's Law (V = IR) is used to find the current (I).
The resistance (R) has a value of 470 ohms with a tolerance of 5%. The current, I = V/R = 7V / 470 ohms = 0.014893617 A, or approximately 14.89 mA.
To specify the current's accuracy, we calculate the uncertainty for each measurement and combine them. The uncertainty for V1 is 0.5V and for V2 is 2%, which is 2% of 5V = 0.1V. The uncertainty in the resistance is 5% of 470 ohms = 23.5 ohms. The total uncertainty in the voltage drop is the sum of V1 and V2's uncertainties, which is 0.5V + 0.1V = 0.6V. The relative uncertainty for the voltage is 0.6V / 7V * 100 = approximately 8.57%. Adding the relative uncertainties, we get 8.57% + 5% = 13.57%. Applying this to the current, the uncertainty is 13.57% of 14.89 mA = approximately 2.02 mA. Therefore, the current is 14.89 mA with an accuracy of ±2.02 mA.