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A battery has a short-circuit current of 30 A and
an open circuit voltage of 24 V. If the battery is connected to an electric bulb of resistance 2Ω, the power dissipated by the bulb is:
a)80 W
b)1800 W
c)146.9 W
d)228 W
Correct ans 'C'. Can you explain this answer?
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To find the current flowing through the bulb when it is connected to the battery, we can use Ohm's Law:

V = I * R

where V is the voltage, I is the current, and R is the resistance.

In this case, the open circuit voltage of the battery is 24 V and the resistance of the bulb is 2 Ω.

We can rearrange the equation to solve for the current:

I = V / R

I = 24 V / 2 Ω

I = 12 A

Therefore, the current flowing through the bulb when it is connected to the battery is 12 A.
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A battery has a short-circuit current of 30 A and ... morean open circuit voltage of 24 V. If the battery is connected to an electric bulb of resistance 2Ω, the power dissipated by the bulb is:a)80 Wb)1800 Wc)146.9 Wd)228 WCorrect ans 'C'. Can you explain this answer?
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A battery has a short-circuit current of 30 A and ... morean open circuit voltage of 24 V. If the battery is connected to an electric bulb of resistance 2Ω, the power dissipated by the bulb is:a)80 Wb)1800 Wc)146.9 Wd)228 WCorrect ans 'C'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A battery has a short-circuit current of 30 A and ... morean open circuit voltage of 24 V. If the battery is connected to an electric bulb of resistance 2Ω, the power dissipated by the bulb is:a)80 Wb)1800 Wc)146.9 Wd)228 WCorrect ans 'C'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A battery has a short-circuit current of 30 A and ... morean open circuit voltage of 24 V. If the battery is connected to an electric bulb of resistance 2Ω, the power dissipated by the bulb is:a)80 Wb)1800 Wc)146.9 Wd)228 WCorrect ans 'C'. Can you explain this answer?.
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