You have a battery marked " 6.00 V ." When you draw a current of 0.361 A from it, the potential difference between its terminals is 5.07 V . What is the potential difference when you draw 0.591 A

Explanation:

Battery voltage is 6V

A current of 0.361A is draw the voltage reduces to, 5.07V

This shows that the appliances resistance that draws the currents is

Using KVL

The battery has an internal resistance r

V=Vr+Va

Vr is internal resistance voltage

Va is appliance voltage

6=5.07+Va

Va=6-5.07

Va=0.93

Using ohms law to the resistance of the appliance

Va=iR

R=Va/i

R=0.93/0.361

R=2.58ohms

Then if the circuit draws a current of 0.591A

Then the voltage across the load is

V=iR

Va=0.591×2.58

Va=1.52V

Then the voltage drop at the internal resistance is

V=Vr+Va

Vr=V-Va

Vr=6-1.52

Vr=4.48V

does anyone know or what?

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