Question

I need step by step calculation!

Ohm's Law A resistor rated at 250 k 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

R = 250 kΩ

V = 3.0 V

R_min = 250*0.95 = 237.5 kΩ

R_max = 250*1.05 = 262.5 kΩ

So I_max = V/R_min = 3/237.5 = 12.6 uA

and I_min = V/R_max = 3/262.5 = 11.4 uA

Answer 2

SOLUTION :

R = 250 kΩ 

Variation possible I resistance value = 5% I.e. - 5% to + 5% .

Hence,

R minimum = 250 * (1 - 0.05) = 237.5 kΩ 

R maximum = 250* (1 + 0.05) = 262.5 kΩ

Voltage applied = 2 * 1,50 = 3.00 V

So, 

Minimum current = 3.00 / (262.5 *10^3) = 11.43 * 10^(-6) Amp = 11.43 µA (ANSWER).

Maximum current = 3.00 / (237.5 *10^3) = 12.633 * 10^(-6) Amp = 12.63 µA (ANSWER).