Question

When a 20.0-ohm resistor is connected across the terminals of a 12.0-V battery, the voltage across the terminals of the battery falls by 0.300 V. What is the internal resistance of this battery

Answer 1

Answer:

The  internal resistance is  $r = 0.5 \ \Omega$

Explanation:

From the question we are told that the resistance of

   The  resistance of the resistor is  $R = 20.0\ \Omega$

    The  voltage is $V = 12.0 \ V$

     The magnitude of the voltage fall is  $e = 0.300\ V$

Generally the current flowing through the terminal due to the voltage of the battery  is  mathematically represented as

        $I = \frac{V}{R}$

substituting values

        $I = \frac{12.0 }{20 }$

       $I = 0.6 \ A$

The internal resistance of the battery is mathematically represented as

      $r = \frac{e}{I}$

substituting values

     $r = \frac{0.300}{ 0.6 }$

    $r = 0.5 \ \Omega$

Answer 2

The internal resistance of the battery is 0.5 ohms.

To calculate the internal resistance of the battery, we use the formula below

Formula:

• (V/R)r = V'............. Equation 1

Where:

• V = Voltage across the terminal of the battery
• R = Resistance connected across the battery
• r = internal resistance of the battery
• V' = voltage drop of the battery.

Make r the subject of the equation

• r = V'R/V............ Equation 2

From the question,

Given:

• V = 12 V
• R = 20 ohms
• V' = 0.3 V

Substitute these values into equation 2

• r = (0.3×20)/12
• r = 6/12
• r = 0.5 ohms.

Hence, The internal resistance of the battery is 0.5 ohms.

Learn more about internal resistance here: https://brainly.com/question/14883923