Chapter 22, Problem 102A

To determine

Rank the resistors from greatest to least on the basis of current.

Answer to Problem 102A

  $E\left(10\text{M}\Omega \right)>B\left(15\text{M}\Omega \right)\text{>}C\left(15\text{M}\Omega \right)>A\left(10\text{M}\Omega \right)>D\left(20\text{k}\Omega \right)$

Explanation of Solution

Given:

A: $R=10\text{M}\Omega V=1.5\text{V}$

B: $R=10\text{M}\Omega V=3\text{V}$

C: $R=15\text{M}\Omega V=1.5\text{V}$

D: $R=15\text{M}\Omega V=0.75\text{V}$

E: $R=20\text{k}\Omega V=9\text{V}$

Formula Used:

From Ohm’s law,

  $V=IR$

Where, $V$ is voltage, $I$ is current and $R$ is resistance.

Calculation:

From above relation current is

  $I=\frac{V}{R}$

For A:

  $I=\frac{1.5}{10\times {10}^6}=0.15\text{μA}$

For B:

  $I=\frac{3}{10\times {10}^6}=0.3\text{μA}$

For C:

  $I=\frac{1.5}{15\times {10}^6}=0.1\text{μA}$

For D:

  $I=\frac{0.75}{15\times {10}^6}=0.05\text{μA}$

For E:

  $I=\frac{9}{20\times {10}^3}=0.45\text{mA}$

The maximum current is across the $20\text{k}\Omega$ resistor.

Conclusion:

Therefore, the resistor ranking based on current value is,

E > B > C > A > D