Three resistors having resistances of R1 = 1.52?, R2 = 2.27?, and R3 = 4.75? res

Question

Three resistors having resistances of R1 = 1.52?, R2 = 2.27?, and R3 = 4.75? respectively, are connected in series to a 27.5V battery that has negligible internal resistance.

Part A.Find the equivalent resistance of the combination. Req =

Part B. Find the current in each resistor. IR1, IR2, IR3 =

Part C. Find the total current through the battery. I =

Part D. Find the voltage across each resistor. VR1, VR2, VR3 =

Part E. Find the power dissipated in each resistor. PR1, PR2, PR3 =

Part F. Which resistor dissipates the most power, the one with the greatest resistance or the one with the least resistance? Explain.

Explanation / Answer

req=r1+r2+r3

=1.52+2.27+4.75

=8.54

I=v/req

=27.5/8.54

=3.22 amp

resistor have same current bcoz they r connected in series

volatage

vr1=3.22*1.52=4.89

vr2=2.27*1.52=3.45

vr3=4.75*1.52=7.22

power

pr1=1.52*4.89=7.43

pr2=1.52*3.45=5.24

pr3=1.52*7.22=10.97

most resistane ..power is directly propertional to resistance

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