Question

The open-circuit voltage of a standard car battery is 12.6 V and the short-circuit

current is approx. 300A. What is the available power from the battery?​

Answer 1

Given:-

• Potential difference ,V = 12.6 V

• Current ,I = 300 A

To Find:-

• Power ,P

Solution:-

By using this Formula

• Power = Potential Difference × Current

→ P = VI

Substitute the value we get

→ P = 300×12.6

→ P = 3780 watt

→ P = 3780/1000 = 3.78 Kj

Therefore the power available from the battery is 3.78 Kilowatt.

Additional Information!!

Potential Difference: The potential difference between two point in an electric circuit carrying some current as the work done to move a unit charge from one point to the other.

Potential difference = Work done /Charge.

V = W/Q

here

V is the potential difference

W is the work done

Q is the charge

✧ SI unit of electric potential energy is volt (V).

✧ 1 volt : The potential difference between two points in a current carrying conductor when one joule of work is done to move a charge of one coulomb from one point to other.

1 Volt = 1Joule/1Coulomb

Answer 2

Answer:

✯ Given :-

• The open-circut voltage of a standard car battery is 12.6 V and the short-circut current is approx. 300 A.

✯ To Find :-

• What is the available power from the battery.

✯ Formula Used :-

$\boxed{\bold{\small{✮\: Power\: =\: Potential\: difference\: ×\: Current}}}$

✯ Solution :-

$\leadsto\sf Given \begin{cases} & \sf{Potential\: difference\: (V)\: = \bf{12.6\;V}} \\ & \sf{Current\: (I) = \bf{300\;A}} \end{cases}$

➣ According to the question by using the formula we get,

✭ P = VI ✭

⇒ P = 12.6 × 300

⇒ P = 3780 watt

⇒ P = $\dfrac{3780}{1000}$

➠ P = 3.78 kw

$\therefore$ The available power from the battery is 3.78 kw .

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