
A secondary cell after long use has an e.m.f of 1.9V and a large internal resistance of $380\Omega $. What maximum current can be drawn from the cell. Could the cell drive the starting motor of a car:
A) 5m, yes.
B) < 5mA, yes.
C) 5mA, no.
D) < 5mA, no.
Answer
566.4k+ views
Hint:A resistance is opposition to the flow of current in an electrical circuit. The internal resistance is the resistance offered by the battery itself to the opposition of the current by the battery which results in heat generation.
Formula used:
The formula of the current for internal resistance in a battery is given by,
${I_{\max .}} = \dfrac{E}{r}$
Where maximum current is ${I_{\max .}}$ the e.m.f of the battery is $E$ and the internal resistance is r.
Complete step by step answer:
A secondary cell after long use has an e.m.f of 1.9V and a large internal resistance of $380\Omega $ we need to find the maximum current draw and also predict whether this cell could start the motor of a car or not.
Let us calculate the maximum current that can be drawn from the secondary cell having an e.m.f of 1.9V and the internal resistance of $380\Omega $.
The formula of the current for internal resistance in a battery is given by,
${I_{\max .}} = \dfrac{E}{r}$
Where maximum current is ${I_{\max .}}$ the e.m.f of the battery is $E$ and the internal resistance is r.
$ \Rightarrow {I_{\max .}} = \dfrac{E}{r}$
As the e.m.f is 1.9V and the internal resistance is $380\Omega $ therefore the maximum current is equal to,
$ \Rightarrow {I_{\max .}} = \dfrac{{1 \cdot 9}}{{380}}$
$ \Rightarrow {I_{\max .}} = 0 \cdot 005A$.
The maximum current from the secondary cell is equal to${I_{\max .}} = 0 \cdot 005A$ and as the maximum current is very less therefore the secondary cell cannot drive a starting motor of a car.
The correct answer for this problem is option C.
Note:The resistance of the circuit in which the battery is joined is different and the internal resistance is different. Students should not get confused in this part also when the total resistance is calculated then the internal resistance is also taken into account. The equivalent resistance is taken out on the basis of the resistance on the circuit and the battery is connected in series or parallel.
Formula used:
The formula of the current for internal resistance in a battery is given by,
${I_{\max .}} = \dfrac{E}{r}$
Where maximum current is ${I_{\max .}}$ the e.m.f of the battery is $E$ and the internal resistance is r.
Complete step by step answer:
A secondary cell after long use has an e.m.f of 1.9V and a large internal resistance of $380\Omega $ we need to find the maximum current draw and also predict whether this cell could start the motor of a car or not.
Let us calculate the maximum current that can be drawn from the secondary cell having an e.m.f of 1.9V and the internal resistance of $380\Omega $.
The formula of the current for internal resistance in a battery is given by,
${I_{\max .}} = \dfrac{E}{r}$
Where maximum current is ${I_{\max .}}$ the e.m.f of the battery is $E$ and the internal resistance is r.
$ \Rightarrow {I_{\max .}} = \dfrac{E}{r}$
As the e.m.f is 1.9V and the internal resistance is $380\Omega $ therefore the maximum current is equal to,
$ \Rightarrow {I_{\max .}} = \dfrac{{1 \cdot 9}}{{380}}$
$ \Rightarrow {I_{\max .}} = 0 \cdot 005A$.
The maximum current from the secondary cell is equal to${I_{\max .}} = 0 \cdot 005A$ and as the maximum current is very less therefore the secondary cell cannot drive a starting motor of a car.
The correct answer for this problem is option C.
Note:The resistance of the circuit in which the battery is joined is different and the internal resistance is different. Students should not get confused in this part also when the total resistance is calculated then the internal resistance is also taken into account. The equivalent resistance is taken out on the basis of the resistance on the circuit and the battery is connected in series or parallel.
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