Assuming we can apply Ohm's Law to the circuit, as the internal resistance and the load resistor are in series, we can find the current I₁ as follows:

$I_{1} = \frac{V}{R_{int} +r_{L} }$

$I_{1} = \frac{V}{R_{int} +r_{L} } = \frac{V}{2*r} (1)$

When the battery ages, if the internal resistance triples, the new current can be found using Ohm's Law again:

$I_{2} = \frac{V}{R_{int} +r_{L} } = \frac{V}{(3*r) +r} = \frac{V}{4*r} (2)$

We can find the relationship between I₂, and I₁, dividing both sides, as follows:

$\frac{I_{2} }{I_{1} } = \frac{V}{4*r} *\frac{2*r}{V} = \frac{1}{2}$

The current when the internal resistance triples, is half of the original value, when the internal resistance was r, equal to the resistance of the load.

7 0

3 years ago

Talja [164]

Be heavier

density=mass÷volume

if two items have the same size they have the same volume so the heavier one will be the denser one

6 0

3 years ago

Natali5045456 [20]

Answer:

Frequency is the number of wavelengths, which is measured in hertz.

Explanation:

I have my ways ;)

5 0

3 years ago