Question

A resistor rated at 275k Ω is connected across two D cell batteries (each 1.50 V) in series, with a total voltage of 3.00 V. The manufacturer advertises that their resistors are within 5% of the rated value. What are the possible minimum current and maximum current through the resistor?.

Answer 1

Answer:

The possible minimum current is: $1.04\times10^{-5} A$

the possible maximum current is: $1.14\times10^{-5} A$

Explanation:

In this circuit the voltage across the resistor is the voltage the batteries provide it (3.00 V), due Ohm's law current (I), voltage (V) and resistance are related by:

$V=IR$

so current is:

$I=\frac{V}{R}$ (1)

Now if we know a resistor is rated are 275k but with 5% of uncertainty, the real value is between:

$275k \ohm - (275k \ohm * 0.05) = 261.25 k\ohm$Ω

and

$275k \ohm + (275k \ohm * 0.05) = 288.75 k\ohm$Ω

This is between 261.25kΩ and 288.75kΩ, those are the minimum and maximum possible values of the resistance respectively. So, using those values on (1) we can find maximum and minimum possible value of current:

$I_{max}=\frac{3.00}{261250}=1.14\times10^{-5} A$

$I_{min}=\frac{3.00}{288750}=1.04\times10^{-5} A$