Question
I =
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 6n 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?
The correct answer is: 20
Hint:-
We will substitute the values given of I, V & R and simplify the equation to find the maximum value of n.
Since the maximum value of I = 0.25 amperes, the corresponding value of n would also give us the maximum Potential difference that can be produced by n batteries and hence, the greatest number of batteries that can be used.
Step-by-step solution:-
I = current = 0.25 ampere
V = Patential Difference = 6n volts
R = Resistance = 500 ohms
I =
∴ 0.25 =
∴ 0.25 × 500 = 6n
∴ 125 = 6n
∴ = n
∴ n ≈ 20.83
Since the maximum value of n is 20.83, we cannot round it up to 21 as that would be exceeding the maximum value.
∴ we round it down to 20.
Final Answer:-
∴ The greatest number of batteries that can be used to have a current (I) of no more than 0.25 amperes with a Resistance (R) of 500 ohms is 20 that will produce a Petential difference (V) of 6n i.e. 6 (20) i.e. 120 volts.
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