A battery has an internal resistance of 1.3 $\varOmega$ and a voltage of 48 V.What is the load resistance which causes the voltage to drop to 45.4 V?

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Quizplus · Accepted Answer

The voltage drop across the internal resistance is the difference between the battery voltage and the load voltage, which is $48V - 45.4V = 2.6V$. The current through the circuit can be found using Ohm's law $I = \frac{V}{R}$, where $V$ is the voltage across the internal resistance and $R$ is the internal resistance. Thus, $I = \frac{2.6V}{1.3\Omega} = 2A$. The voltage across the load is 45.4V, so using Ohm's law again, the load resistance $R_{load} = \frac{V_{load}}{I} = \frac{45.4V}{2A} = 22.7\Omega$.

A Battery Has an Internal Resistance of 1 Ω\varOmegaΩ And a Voltage of 48 V

A Battery Has an Internal Resistance of 1 $\varOmega$ And a Voltage of 48 V