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'Battery Modelling' printed from https://nrich.maths.org/
Part 1:
The ammeter will drop a voltage of 2V at 20A, it can therefore be
modelled as a resistor of 0.1$\Omega$.
The $100\Omega$ resistor is in parallel with the series combination
of the ammeter and battery.
$100 I_1 = (0.1x2) + (2x2) + 24$
$I_1 = 0.282 A$
Current Conservation:
$I_0 = I_1 + I_2$
$I_0 = 2 + 0.282 = 2.282 A$
Part 2:
Power supplied = Power consumed
Power Supplied = $ (0.282^2 x100) + (2^2 x 0.1) + (2x2^2) + (2x24)$
= 64.35 W
Power dissipated by battery = IV = 2 x 24 = 48W
74.6 % of the power supllied is fed into the battery.
Part 3:
$I_0$ is unchanged, $I_0$ = 2.282A
Current conservation:
$I_0 = I_1 + I_2$
Voltage Conservation:
$100 I_1 = 24 + 2I_2 $
Solving these two equations simultaneously we find $I_2 = \frac{I_0
- 0.24}{1.02} = 2.002 A$