The performance of the starter circuit in an automobile can be significantly degraded by a small amount of corrosion on a battery terminal. The figure (a) depicts a properly functioning circuit with a battery (12.5 V emf, 0.02 N internal resistance) attached via corrosion-free cables to a starter motor of resistance Rs = 0.15 N. Suppose that later, corrosion between a battery terminal and a starter cable introduces an extra series resistance of just Rc figure (b). Let Po be the power delivered to the starter in the circuit free of corrosion, and let P be the power 0.10 N into the circuit as suggested in the delivered to the circuit with corrosion.

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26.17

Determine the ratio P/P0

The performance of the starter circuit in an automobile
can be significantly degraded by a small amount of
corrosion on a battery terminal. The figure (a) depicts a
properly functioning circuit with a battery (12.5 V emf,
0.02 2 internal resistance) attached via corrosion-free
cables to a starter motor of resistance Rs = 0.15 N.
Suppose that later, corrosion between a battery terminal
and a starter cable introduces an extra series resistance
of just Rc = 0.10 N into the circuit as suggested in the
figure (b). Let Po be the power delivered to the starter in
the circuit free of corrosion, and let P be the power
delivered to the circuit with corrosion.
Transcribed Image Text:The performance of the starter circuit in an automobile can be significantly degraded by a small amount of corrosion on a battery terminal. The figure (a) depicts a properly functioning circuit with a battery (12.5 V emf, 0.02 2 internal resistance) attached via corrosion-free cables to a starter motor of resistance Rs = 0.15 N. Suppose that later, corrosion between a battery terminal and a starter cable introduces an extra series resistance of just Rc = 0.10 N into the circuit as suggested in the figure (b). Let Po be the power delivered to the starter in the circuit free of corrosion, and let P be the power delivered to the circuit with corrosion.
& = 12.5 V
r=0.02 2
ww-
Rg =0.15 2
(a)
Rc=0.10 2
r0.02 2
ww
Rg = 0.15 2
(b)
Transcribed Image Text:& = 12.5 V r=0.02 2 ww- Rg =0.15 2 (a) Rc=0.10 2 r0.02 2 ww Rg = 0.15 2 (b)
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