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anonymous

  • 5 years ago

The current in a series circuit is 1 amp. What will the current be when the resistance is increased by 75%

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  1. anonymous
    • 5 years ago
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    Ohm's Law is V=IR. Rearranging, you have I = V/R. Now, assuming your voltage remains constant when the new resistance is installed, you will have\[\frac{I_2}{I_1}=\frac{V/R_2}{V/R_1}=\frac{R_1}{R_2}\]So your new current is,\[I_2=(\frac{R_1}{R_2})I_1=\frac{R_1}{1.75R_1}1A=\frac{4}{7}A\]

  2. anonymous
    • 5 years ago
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    thank you so much

  3. anonymous
    • 5 years ago
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    No worries.

  4. anonymous
    • 5 years ago
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    I have a question do you tutor or are you a student

  5. anonymous
    • 5 years ago
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    Hi there, I'm not a student, or a tutor. I have a degree in mathematics and physics.

  6. anonymous
    • 5 years ago
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    A 4.7 kΩ resistor dissipates 0.75 W. The voltage is can you help me with this

  7. anonymous
    • 5 years ago
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    You know have resistance and power, and you need to find voltage.  When I see 'resistance' in a question, I tend to think, V=IR When I see power, I tend to think,  P=IV These equations relate the thing you want - V - to variables you have - P and R.  The one thing they have in common that we don't want, is current, I.  So why don't we solve the first equation for I and substitute it into the second? So, I=V/R from first eq'n, and into the second,  P=(V/R)*V=V^2/R Solving for V, V = sqrt(PR) = sqrt(.75W*4700(ohms))= sqrt(3525) volts ~ 59V Check the actual arithmetic - I'm doing this on an iPhone and it's difficult. 

  8. anonymous
    • 5 years ago
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    Im sorry this was the wrong problem I don';t know why it went through twice

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