Chapter 29.2, Problem 26PP

To determine

To calculate the potential difference of the battery to produce a current of 2.5 mA in a diode.

Answer to Problem 26PP

The potential difference of the battery is 1.875 V.

Explanation of Solution

Given data:

Current ( I ) = 25 mA= 0.0025 A

  $V_{diode}=0.70$ V

Resistance ( R ) = 470 ohm

Formula used:

Voltage required ( V ),

  $V=IR+V_{diode}$

Calculation:

Current ( I ) = 25 mA= 0.0025 A

Voltage required ( V ),

  $V=IR+V_{diode}=\left(0.0025\right)\left(470\right)+0.70=1.875\text{ V}$

Conclusion:

Hence, the potential difference of the battery is 1.875 V.