A real battery is not just an emf. We can model a real 1.5 V battery as a 1.5 V

Question

A real battery is not just an emf. We can model a real 1.5 V battery as a 1.5 V emf in series with a resistor known as the "internal resistance", as shown in the figure(Figure 1) . A typical battery has 1.0 internal resistance due to imperfections that limit current through the battery. When there's no current through the battery, and thus no voltage drop across the internal resistance, the potential difference between its terminals is 1.5 V, the value of the emf. Suppose the terminals of this battery are connected to a 2.1 resistor. What is the potential difference between the terminals of the battery? What fraction of the battery's power is dissipated by the internal resistance?

Explanation / Answer

Given that

emf (e) =1.5V

Intenal resitance(r) =1.0ohm

The potential difference between the erminal (V) =1.5V

The resistance connected to terminal (R) =2.1ohm

We consider here e is the emf of the battery , r be the internal resistance and R be the external resistance to which the battery is connected then

V = eR/(R+r) =1.5*2.1/(2.1+1)=3.15/3.1=1.016V

Now the power dissipiated through the internal resistance is given by

P =I2*r

now the circuit in the circuit is given by

I =e/(R+r) =1.5/3.1 =0.483A

Now power P =(0.483)2*(1.0)=0.233W

Now the total power dissipated from both the internal and external resistances are

P =(0.483)2*(3.1) =0.7231W

Now the percentage of power dissipiated by the battery is (0.7231)/0.233 =3.103%

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