Current provided by the battery is maximum when ?
Answer
588.3k+ views
Hint:In a general condition, the current can be defined from the Ohm’s Law as the ratio of the total voltage supplied to the circuit to the total resistance offered by the circuit. The total resistance includes the internal resistance of the battery as well as the external resistance. As resistance is inversely proportional to the current for a constant voltage, for maximum current the resistance needs to be minimum.
Complete step by step answer:
Here, we need to find the maximum current $I_{max}$ provided by the battery. Let us assume a battery with internal resistance $r$ that provides emf $E$ to an external resistance $R$ .Hence, the total resistance in the circuit will be the sum of the internal resistance of the battery and the external resistance.
We can infer from the Ohm’s Law ( $V=IR$ ), that the current can be found by dividing the emf of the battery with the total resistance in the circuit which can be expressed mathematically as
$I=\dfrac{E}{r+R}$
Now, we know that the emf of a battery is constant. Hence, from the above equation we can say that the current in the circuit is inversely proportional to the total resistance of the circuit. Hence, as we need to find the maximum current here, we will need the total resistance to be minimum.
Now, the total resistance is ${{R}_{t}}=r+R$
We can rewrite the total resistance as
${{R}_{t}}={{\left( \sqrt{r} \right)}^{2}}+{{\left( \sqrt{R} \right)}^{2}}$
The middle term required to make the above equation a perfect square would be $2\sqrt{r}\sqrt{R}$
$\Rightarrow {{R}_{t}}={{\left( \sqrt{r} \right)}^{2}}+{{\left( \sqrt{R} \right)}^{2}}+2\sqrt{r}\sqrt{R}-2\sqrt{r}\sqrt{R}$
$\Rightarrow {{R}_{t}}={{\left( \sqrt{r} \right)}^{2}}-2\sqrt{r}\sqrt{R}+{{\left( \sqrt{R} \right)}^{2}}+2\sqrt{r}\sqrt{R}$
Now, we can write the above equation in the terms of a perfect square as,
${{R}_{t}}={{\left( \sqrt{r}-\sqrt{R} \right)}^{2}}+2\sqrt{r}\sqrt{R}$
We know that we require the minimum value of the resistance. The minimum value from the above equation will be obtained when the squared term will be equal to zero.
$\sqrt{r}-\sqrt{R}=0$
$\Rightarrow \sqrt{r}=\sqrt{R}$
Squaring on both sides,
$\therefore r=R$
Hence, when the external resistance will be equal to the internal resistance of the battery, the total resistance will be minimum and the current will be maximum.
Note: We can understand from the Ohm’s Law that the current can reach to infinity if the resistance is zero (which is the condition for superconductors). However, this can only be achieved hypothetically. In practical applications, a battery always has an internal resistance which cannot be eliminated. Also the wire used in the connection of the circuit has a small amount of resistance, but it can be neglected as compared to the external resistance.
Complete step by step answer:
Here, we need to find the maximum current $I_{max}$ provided by the battery. Let us assume a battery with internal resistance $r$ that provides emf $E$ to an external resistance $R$ .Hence, the total resistance in the circuit will be the sum of the internal resistance of the battery and the external resistance.
We can infer from the Ohm’s Law ( $V=IR$ ), that the current can be found by dividing the emf of the battery with the total resistance in the circuit which can be expressed mathematically as
$I=\dfrac{E}{r+R}$
Now, we know that the emf of a battery is constant. Hence, from the above equation we can say that the current in the circuit is inversely proportional to the total resistance of the circuit. Hence, as we need to find the maximum current here, we will need the total resistance to be minimum.
Now, the total resistance is ${{R}_{t}}=r+R$
We can rewrite the total resistance as
${{R}_{t}}={{\left( \sqrt{r} \right)}^{2}}+{{\left( \sqrt{R} \right)}^{2}}$
The middle term required to make the above equation a perfect square would be $2\sqrt{r}\sqrt{R}$
$\Rightarrow {{R}_{t}}={{\left( \sqrt{r} \right)}^{2}}+{{\left( \sqrt{R} \right)}^{2}}+2\sqrt{r}\sqrt{R}-2\sqrt{r}\sqrt{R}$
$\Rightarrow {{R}_{t}}={{\left( \sqrt{r} \right)}^{2}}-2\sqrt{r}\sqrt{R}+{{\left( \sqrt{R} \right)}^{2}}+2\sqrt{r}\sqrt{R}$
Now, we can write the above equation in the terms of a perfect square as,
${{R}_{t}}={{\left( \sqrt{r}-\sqrt{R} \right)}^{2}}+2\sqrt{r}\sqrt{R}$
We know that we require the minimum value of the resistance. The minimum value from the above equation will be obtained when the squared term will be equal to zero.
$\sqrt{r}-\sqrt{R}=0$
$\Rightarrow \sqrt{r}=\sqrt{R}$
Squaring on both sides,
$\therefore r=R$
Hence, when the external resistance will be equal to the internal resistance of the battery, the total resistance will be minimum and the current will be maximum.
Note: We can understand from the Ohm’s Law that the current can reach to infinity if the resistance is zero (which is the condition for superconductors). However, this can only be achieved hypothetically. In practical applications, a battery always has an internal resistance which cannot be eliminated. Also the wire used in the connection of the circuit has a small amount of resistance, but it can be neglected as compared to the external resistance.
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