2/23/2014 Master Ithttp://www.webassign.net/v4cgi/questions/tutorial_popup.tpl 1/4Master ItThe circuit shown in the figure below is connected for 2.60 min. (Assume R1 = 8.25 Ω, R2 = 2.05 Ω, and V =13.5 V.)(a) Determine the current in each branch of the circuit.(b) Find the energy delivered by each battery.(c) Find the energy delivered to each resistor.(d) Identify the type of energy storage transformation that occurs in the operation of the circuit.(e) Find the total amount of energy transformed into internal energy in the resistors.Part 1 of 10 ConceptualizeWe estimate a current of about 1 A in the 13.5V battery, in the 8.25Ω resistor, and a much smaller currentin the 4.00V battery. The total energy conversion might be about Part 2 of 10 CategorizeFirst we name the currents as shown in the diagram. Then we apply Kirchhoff's rules to current and voltageand solve the linear system of equations to find the resulting currents. For the energies delivered by eachbattery, we multiply voltage times current over time;; for the energies delivered to each resistor, we multiplythe square of the current times resistance over time. To find the total amount of energy transformed, we addup the energies either delivered by the batteries or delivered to the resistors.Part 3 of 10 AnalyzeFrom Kirchhoff's junction rule and by Kirchhoff's loop rule, Clockwise aroundthe lefthand loop containing I1 and I2, we have(27 J/s)(2.60 min)(60 s/min) = 4 212 J.I = 0,junctionΔV = 0.closed loop+I1(8.25 Ω) − I2(5.00 Ω) − I2(1.00 Ω) − 4.00 V = 0.2/23/2014 Master Ithttp://www.webassign.net/v4cgi/questions/tutorial_popup.tpl 2/4This becomesApplying Kirchhoff's loop rule clockwise around the righthand loop containing I2 and I3 and combiningresistors and voltage, givesTo solve three independent equations in three unknowns, we substitute for I3 and reduce the threeequations to two equations as follows.Solving the first equation for I2 givesRearranging the second of the pair of equations givesWe substitute for I2 to obtain one equation in one unknown. Then we solve for the current I1 down the 8.25Ωresistor. We haveNow we can substitute the value we calculated for I1 to find the current down the middle branchand the current in the righthand loopPart 4 of 10 Analyze(a) Summarizing our results, we have .833 0.833 A down the 8.25Ω resistor, .479 0.479 A down the middle branch, and 1.31 1.31 A up in the righthand branch.Part 5 of 10 Analyze(b) The power converted by a battery is so the energy delivered by the 4.00V battery isgiven by the following equation. Remember: the sign of the current is positive when it is in the direction ofincreasing potential and negative when it is in the direction of decreasing potential for the battery.For the energy delivered by the 13.5V battery, we havePart 6 of 10 Analyze(c) For a resistor, so that the energy converted becomes For the energy delivered tothe resistors, we have the following:(8.25 Ω)I1 − (6.00 Ω)I2 − 4.00 V = 0.(6.00 Ω)I2 + (5.05 Ω)I3 − 9.50 V = 0.(I1 + I2)(8.25 Ω)I1 − (6.00 Ω)I2 − 4.00 V = 0(5.05 Ω)(I1 + I2) + (6.00 Ω)I2 − 9.50 V = 0I2 = .(8.25 Ω)I1 − 4.00 V6.00 ΩI1 + I2 = .11.05 Ω5.059.50 V5.05I1 = 0.833 0.833 V/Ω = .833 0.833 A.I2 = = .4789 0.479 A8.25 Ω .833 0.833 A − 4.00 V6.00 ΩI3 = I1 + I2 = 1.312 1.31 A.P = ΔVI = U/Δt,U = 4.00 V -‐‑.479 0.479 A 156 s = -‐‑298 299 J.U = 13.5 V 1.31 1.31 A 156 s = 2758 2760 J.ΔV = IR, U = I2RΔt.22/23/2014 Master Ithttp://www.webassign.net/v4cgi/questions/tutorial_popup.tpl 3/4to the resistor: to the 5.00Ω resistor: Part 7 of 10 AnalyzeFor the energy delivered to the remaining resistors, we have the following:to the 1.00Ω resistor in the center branch:to the 3.00Ω resistor: and to the resistor in the righthand branch:Part 8 of 10 Analyze(d) As the circuit carries current, the 4.00-‐‑V 13.5V battery converts chemical energy intoelectrically transmitted energy. The 13.5-‐‑V 4.00V battery converts 299 Jfrom electrically transmitted energy into chemical energy.Part 9 of 10 AnalyzeAll of the -‐‑-‐‑-‐‑Select-‐‑-‐‑-‐‑ resistors convert electrically transmitted energy into internal energy. The nettransformation, in terms of energy, is from -‐‑-‐‑-‐‑Select-‐‑-‐‑-‐‑ chemical energy into -‐‑-‐‑-‐‑Select-‐‑-‐‑-‐‑ internal energy.Part 10 of 10 Analyze(e) The total amount of energy transformed to internal energy in the resistors is found by adding the energiesdelivered to each resistor. We haveThe total internal energy in the resistors is -‐‑-‐‑-‐‑Select-‐‑-‐‑-‐‑ equal to
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