Question

5.00 Ω 3.00 Ω W R2 1.00 Ω R, + + 4.00 V -Τ.

The circuit shown in the figure below is connected for 2.70 min. (Assume R₁ = 8.30 Ω, R₂ = 2.70 Ω, and V = 11.0 V.)

(a) Determine the current in each branch of the circuit. (left branch, middle branch, right branch)

(b) Find the energy delivered by each battery.

4.00 V battery	J
11.0 V battery	kJ

(c) Find the energy delivered to each resistor.

8.30 Ω resistor	J
5.00 Ω resistor	J
1.00 Ω resistor	J
3.00 Ω resistor	J
2.70 Ω resistor	J

(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.

Answer 1

352 R2 RGin +. V gestad toov 4 4 6 I 28.3I - 8. 3 I to Solution:- time = 2.7 min 32 (ITI R = 8.32 R₂ = 2.700 I 7 Iz V = Il Volt 4.00 V I. Part :- Now using kirchoff's law to determine the current in current in each branchino for loop 4 : V-I, R2 = 3 I, +5 I2 + 1 I 2 - 4 =0 11- I (2.70)-31, +6 12-4 = 0 IS FERO 75.7 I, + 6 Iz orord albbins Losoud 5.7.7 - 6 I2 = 7 for loop Li- 4-112-512-R (7, +12) = 0 ai tosiu) 02 (ث) x 8.31, رنت - 4-14.3 12 8:31, 30 14.3 I2 +8.31, -4 from eq ci- bocu.200.0 12) = 4 5.7 pe 14.312 +10.193 +8.73612-4 SOLYA 23.036 6 Iz= 6· 193 (odyt 14.3 12+8;3171 5.7 T he 12=-0.268 A

Put Iz in eqli) - 571-66-0.260)=7 SEB 5,7% = 5.387 toimit 1 I = 0,945 A scicarotaba na and an med (I, +12) = 0.945 -0.268 A-s 1 (1+1) = 0.677) I E-(of S), So- Current in left branch = 0.677 A middle branch = 0.268 A Current in right branch = 0.945 A et go Current in Ans Part(b) serta sta Power delivered by V volt battery ī vx It = tex0.945 = 10.395 Watt ci So- energy (E) Pxt 10.395 x(2.7x60) TE 2 TET.+ Evvalt I 1683.99 J 1.684x 103 J Ev =10684 kJ Am

Ea vost Am Energy delivered by 8.35 resistor- Energy delivered by delivered by 4.00 volt by 4.00 volt battery forgo Eaveet = Pxto (4x0.268 X 2.7x60 ) I 173,664 J Part (C) & to tovaras lolot E334 = Pxt = [(1+1))}<R] + = [(0.667) 48. 3) x(2.7x60) E8.32 = 607 0166] Energy delivered by 5.000 resistor- Esos [ 12x5 x 2.7x60] = ((0.268)*5x2.7 x60] Esor = 58.177 ) An delivered by for resistor En = (0.268)* 1x2.7x60 11.635 J Am Am Energy Eur Energy delivered by 3.00r resistor- Ezr = (0.945), x3x 2.7x60 Ear = 434.013 delivered by 2.78 resistor E2.72 =10.945)x2.7 x 60X2:7 Am Am Energy E2.70 = 390.609 J

Part(d);- The energy will be in the of heat energy in the operation of the circuit Am Partie); Total amount of energy transformed into internal energy in the resistor- E- 1501.598 Joule Am