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

Question

tino4ka555 [31]

4 years ago

7

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

Physics

2 answers:

AfilCa [17] · Answer 1 · 2022-01-15T17:24:04+03:00

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.