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I need help with physics!!!

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7b)
• The Voltage is fixed since it is determined by the rate of change of flux.
• Higher resistance, means less Induced Current since R=V/I.
• Less induced current means less energy loss by motor effect, therefore less damping. P = V x I or P = V^2 / r but you cant use P = I^2 R since the current changes x2!
--- > If you do this in an open circuit. Resistance is maximum , current is zero, so no damping.

ci)
Above, not zero :¬
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Help me out with this :/
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oh thnk god im still sane :3
You saw that answer ? If not then :¬
7b)
• The Voltage is fixed since it is determined by the rate of change of flux.
• Higher resistance, means less Induced Current since R=V/I.
• Less induced current means less energy loss by motor effect, therefore less damping. P = V x I or P = V^2 / r but you cant use P = I^2 R since the current changes x2!
--- > If you do this in an open circuit. Resistance is maximum , current is zero, so no damping.

ci)
Above, not zero :¬
View attachment 45786
 
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yea i saw it.i meant im still sane coz it was right ^_^

btw u shldve drawn the graph on fig 7.2....thts wht im actually confused abt -_-
You said for c question. Dont lie.
for 7.2 clue,
• The Voltage is fixed since it is determined by the rate of change of flux.
• Higher resistance, means less Induced Current since R=V/I.
• Less induced current means less energy loss by motor effect, therefore less damping. P = V x I or P = V^2 / r but you cant use P = I^2 R since the current changes x2!
--- > If you do this in an open circuit. Resistance is maximum , current is zero, so no damping.
 
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You said for c question. Dont lie.
for 7.2 clue,
• The Voltage is fixed since it is determined by the rate of change of flux.
• Higher resistance, means less Induced Current since R=V/I.
• Less induced current means less energy loss by motor effect, therefore less damping. P = V x I or P = V^2 / r but you cant use P = I^2 R since the current changes x2!
--- > If you do this in an open circuit. Resistance is maximum , current is zero, so no damping.

i was correct abt the dumb question i meant -_-...the ammeter one -_-
 
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