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If the battery has an internal resistance of [tex]5.00 \, \Omega[/tex], what current will the array draw if the battery is connected across BC?

A) \(\frac{5.00 \, A}{R_{bc}}\)

B) \(\frac{R_{bc}}{5.00 \, A}\)

C) \(\frac{5.00 \, A}{R_{bc} + 5.00 \, \Omega}\)

D) \(\frac{R_{bc} + 5.00 \, \Omega}{5.00 \, A}\)

Answer :

The current drawn by the array in the given circuit is calculated using Ohm's Law, resulting in a final current value of 1.67 A.

Given Data:

Emf = 50.00 V

Equivalent resistance of the circuit = 25.00 Ω

Internal resistance = 5.00 Ω

Calculations:

Calculate the current drawn from the battery using Ohm's Law: I = V / (Req + r)

Substitute the values to find the current: I = 50.00 V / (25.00 Ω + 5.00 Ω) = 1.67 A

Answer: The current drawn by the array when the battery is connected across BC is 1.67 A.

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Rewritten by : Barada

Final answer:

The current drawn by the array with a battery having an internal resistance of 5.00Ω when connected across BC is calculated by dividing the voltage across BC by the sum of the external resistance and the internal resistance of the battery. Therefore, the correct option is C) 5.00A/Rbc + 5.00Ω.

Explanation:

When considering a circuit where a battery with an internal resistance of 5.00Ω is connected across terminals B and C, the current drawn by the array can be determined by Ohm's Law, which states I = V / (R + r), where I is the current, V is the voltage, R is the external resistance, and r is the internal resistance of the battery. The question presents a formula without specifying the voltage or the resistance across BC (Rbc), and thus we cannot give a numeric answer. However, to find the current, you would divide the voltage across BC by the sum of Rbc and the internal resistance (5.00Ω).

If Rbc is known, the correct expression for the current would be I = V / (Rbc + 5.00Ω), where V is the voltage across BC. None of the options A) through D) provided in the question are typically correct without context or specific numbers, but option C) resembles the form of the Ohm's Law equation if Rbc is replaced with the voltage across BC and the plus sign is replaced with the division sign.