Question

Suppose a car battery has an internal resistance of 0.02 ohms . The voltage between terminals when no current is flowing is 13.6 volts. (The usual value of "12 volts" for lead-acid batteries is only nominal; the actual voltage is higher.) When the battery is supplying 80 A of current to a starter motor, what is the voltage between the terminals of the battery? 1 1.6

Answer 1

When the battery is supplying 80A of current, the voltage drop across its internal resistance is

V = (current) · (resistance)

V = (80A) · (0.02 Ω)

V = 1.6 volts

The voltage that appears across the terminals of the battery is

V = (13.6 - 1.6)

V = 12.0 volts

Answer 2

The voltage between the terminals of the car battery is equal to 12 Volts.

Given the following data:

• Internal resistance = 0.02 ohms

• Voltage (no current) = 13.6 Volts

• Current = 80 Amps

To find the voltage between the terminals of the battery:

The potential difference of a power source when no current is flowing through a circuit is the electromotive force (emf).

Mathematically, the electromotive force (emf) is given by the formula:

$E = V + Ir$

Where:

• I is the current.

• r is the internal resistance.

• V is the voltage.

• E is the electromotive force (emf).

Substituting the given parameters into the formula, we have;

$13.6 = V + 80(0.02)\\\\13.6=V+1.6\\\\V = 13.6 -1.6$

V = 12 Volts

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