The emf and the internal resistance of a battery. If a current of 8.3 A is drawn

Question

The emf and the internal resistance of a battery. If a current of 8.3 A is drawn from the battery when a resistor R is connected across the terminals a=95.0 v, b=5.0 Ohmof the battery, what is the power dissipated by the resistor R? 790 W 440 W 700 W 530 W 620 W The emf and the internal resistance of a battery. If a current of 8.3 A is drawn from the battery when a resistor R is connected across the terminals a=95.0 v, b=5.0 Ohmof the battery, what is the power dissipated by the resistor R? 790 W 440 W 700 W 530 W 620 W 790 W 440 W 700 W 530 W 620 W 790 W 440 W 700 W 530 W 620 W

Explanation / Answer

8.3=95/(R+5) ==>R=6.4458 P=I^2 R =8.3^2 *6.4458 =440

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