Question

A resistor rated at 690 kΩ 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 4% of the rated value. What are the possible minimum current and maximum current through the resistor (in µA)?

Answer 1

Answer:

The possible minimum current and maximum current through the resistor is 4.53 μA and 4.18 μA

Explanation:

Given that,

Resistance = 690 kΩ

Total voltage = 3.00 V

We need to calculate the maximum resistance

$R_{max}=690000+(\dfrac{4}{100}\times690000)$

$R_{max}=717600\ \Omega$

We need to calculate the minimum resistance

$R_{min}=690000-(\dfrac{4}{100}\times690000)$

$R_{min}=662400\ \Omega$

We need to calculate the maximum and minimum current

Using ohm's law

For maximum current,

$V = I R$

$I_{max}=\dfrac{V}{R_{max}}$

$I_{max}=\dfrac{3.00}{717600}$

$I_{max}=0.00000418\ A$

$I_{max}=4.18\times10^{-6}=4.18\ \mu A$

For minimum current,

$I_{max}=\dfrac{3.00}{662400}$

$I_{min}=0.00000453\ A$

$I_{min}=4.53\times10^{-6}= 4.53\ \mu A$

Hence, The possible minimum current and maximum current through the resistor is 4.53 μA and 4.18 μA