Question

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

Answer 1

Answer:

V = 4.48 V

Explanation:

• As the potential difference between the battery terminals, is less than the rated value of the battery, this means that there is some loss in the internal resistance of the battery.

• We can calculate this loss, applying Ohm's law to the internal resistance, as follows:

 $V_{rint} = I* r_{int}$

• The value of the potential difference between the terminals of the battery, is just the voltage of the battery, minus the loss in the internal resistance, as follows:

   $V = V_{b} - V_{rint} = 5.07 V = 6.00 V - 0.361 A * r_{int}$

• We can solve for rint, as follows:

 $r_{int} =\frac{V_{b}- V_{rint} }{I} = \frac{6.00 V - 5.07 V}{0.361A} = 2.58 \Omega$

• When the circuit draws from battery a current I of 0.591A, we can find the potential difference between the terminals of the battery, as follows:

 $V = V_{b} - V_{rint} = 6.00 V - 0.591 A * 2.58 \Omega = 4.48 V$

• As the current draw is larger, the loss in the internal resistance will be larger too, so the potential difference between the terminals of the battery will be lower.  

Answer 2

Answer:

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