HardMath· Algebra· Linear inequalities in one or two variables· SPR
The formula above is Ohm’s law for an electric circuit with current I, in amperes, potential difference V, in volts, and resistance R, in ohms. A circuit has a resistance of 500 ohms, and its potential difference will be generated by n six-volt batteries that produce a total potential difference of volts. If the circuit is to have a current of no more than 0.25 ampere, what is the greatest number, n, of six-volt batteries that can be used?
Correct answer: 20
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Explanation
The correct answer is 20. For the given circuit, the resistance R is 500 ohms, and the total potential difference V generated by n batteries is volts. It’s also given that the circuit is to have a current of no more than 0.25 ampere, which can be expressed as . Since Ohm’s law says that , the given values for V and R can be substituted for I in this inequality, which yields . Multiplying both sides of this inequality by 500 yields , and dividing both sides of this inequality by 6 yields . Since the number of batteries must be a whole number less than 20.833, the greatest number of batteries that can be used in this circuit is 20.
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id cb_question_bank:9c6d3083-e5ee-4c11-a2d6-0e8801f22b3a · question e8f9e117 · IBN 03867-DC
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