R2 Consider the circuit diagram here. Let R1-20 , R2-30 , and R3 = 10 . a) Find
ID: 1629696 • Letter: R
Question
R2 Consider the circuit diagram here. Let R1-20 , R2-30 , and R3 = 10 . a) Find the equivalent resistance of all the resistors combined 1) in this circuit. b) Is the voltage difference supplied by the battery the same magnitude as the voltage drop across any of the individual resistors? If so, which one(s)? e) Is the current passing through the battery the same as the current passing through any of the individual resistors? If so, which one(s)? d) Find the voltage drop across each resistor and the current passing through each resistor. e) Determine the rate at which the battery delivers energy to the circuit. battery What would the voltages across each of the resistors in the preceding circuit be if a wire were placed across the ends of resistor 3 (that is, resistor 3 is shorted out)? Does placing the wire across resistor 3 also short out resistor 2? Explain, using the steady-state energy-density model. 2) 3) Look around you, find an appliance, and look for its power rating. What is the power in watts current does this appliance "draw" if the AC voltage applied to it is 120/? current does this appliance "draw" if the AC voltage applied to it is l2 4) Consider a 60W light bulb plugged into a 120V socket, and a 100W light bulb plugged into another 120V socket. Recall that a bulb's brightness is related to the amount of power (in watts) that is dissipated as heat and light a) Through which light bulb is more current passing? b) Which light bulb filament has more resistance?Explanation / Answer
C) since R1 is in series combination with battery ,then current passing through the battery the same as the current passing through R1
D) equivalent resistance is Req = R1+[(R2*R3)/(R2+R3)] = 20+((30*10)/(30+10)) = 27.5 ohm
net current through the battery is Inet = V/Rnet = 22/27.5 = 0.8 A
current through R1 is i1 = 0.8 A
voltage drop across R1 is V1 = Inet*R1 = 0.8*20 = 16 A
Volatge drop across R2 = R3 = 22-16 = 6 V
current through R2 is i2 = 6/30 = 0.2 A
current through R3 is i3 = 6/10 = 0.6 A
e) Power delivered by battery is P = V*I = 22*0.8 = 17.6 W
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.