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

Question

anonymous · Answer

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,$$I_2=(\frac{R_1}{R_2})I_1=\frac{R_1}{1.75R_1}1A=\frac{4}{7}A$$

anonymous · Answer

thank you so much

anonymous · Answer

No worries.

anonymous · Answer

I have a question do you tutor or are you a student

anonymous · Answer

Hi there, I'm not a student, or a tutor.  I have a degree in mathematics and physics.

anonymous · Answer

A 4.7 kΩ resistor dissipates 0.75 W. The voltage is can you help me with this

anonymous · Answer

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.

anonymous · Answer

Im sorry this was the wrong problem I don';t know why it went through twice