A battery has an internal resistance R_int.Connecting a wire of negligible resistance between the terminals of the battery draws the maximum current I_max from this battery.If instead a resistor is connected between the terminals,a smaller current is produced.What value resistor placed across the terminals would result in the current produced to be half of I_max? A)0.63R_int B)R_int C)0.50R_int D)0.37R_int

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The maximum current $I_{\text{max}}$ is drawn when the external resistance is zero (wire of negligible resistance). To halve this current, the external resistance $R_{\text{ext}}$ must equal the internal resistance $R_{\text{int}}$, as per the formula derived from Ohm's law for circuits with internal resistance: $I = \frac{V}{R_{\text{int}} + R_{\text{ext}}}$. When $R_{\text{ext}} = R_{\text{int}}$, the total resistance doubles, halving the current.

A Battery Has an Internal Resistance Rint

A Battery Has an Internal Resistance R_int