Question

The circuit shown in the figure below is connected for 2.70 min. (Assume R. = 8.80 N, R2 = 2.70 1, and V = 15.0 V.) 5.00 23.00 12 R2 } 1.00 12 R + -4.00 V - direction down (a) Determine the current in each branch of the circuit. branch magnitude (A) left X branch Your response differs from the correct answer by more than 10%. Double check your calculations. x middle Your response is within 10% of the correct value. This may be due to roundoff error, branch or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. right x branch Your response differs from the correct answer by more than 10%. Double check your calculations. down up (b) Find the energy delivered by each battery. 4.00 v battery 15.0 V battery k] ] (c) Find the energy delivered to each resistor. 8.80 resistor 5.00 resistor 1.00 2 resistor 3.00 2 resistor 2.70 resistor ] ] ם (d) Identify the type of energy storage transformation that occurs in the operation of the circuit.

Answer 1

(a) From the current i be the current through left loop and iz be the current through right loop, iz Be the current through Rz Applying Kirchhoff's second law around the left side loop 4-6(i – iz) – 8.8ię = 0 4-6i +6i, – 8.8i = 0 4-14.81 +6i, = 0 -- ---(1) Applying Kirchhoff's second law around the right side loop 15-6(iz – i )-5i2 = 15+61, -1 li, = 0)---(2) From the junction rule, is = iz – į ---(3) By solving the equation, i = 1.05678 A = 1.057 A in = 1.940 A is = 0.883 A (b) The energy delivered by the 4.0 V batteries is, E31 = Vigt = (4.0)(1.05678)(2.70(60)) = 684.79 J The energy delivered by the 11.0 V batteries is, Ex2 =Vint = (15.0)(1.94)(2.70(60)) =4617 J (C) The energy delivered to a 8.802 resistor is, E6.80 = iRt = (1.05678) (8.8)(2.70(60)) =1592.1J The energy delivered to a 5.00 22 resistor is, E5.00 = iz Rt =(0.883)? (5.0)(2.70(60)) = 631.54 J The energy delivered to a 1.00 22 resistor is,

Elon = (iz – 1)? Rt =(0.883)’ (1.0)(2.70 (60)) = 126.31J The energy delivered to a 3.00 22 resistor is, E3.00 = iRt = (1.94) (3.0)(2.70 (60)) =1829.11 J The energy delivered to a 2.70 22 resistor is, E1.600 = iz Rt =(1.94) (2.70)(2.70(60)) =1646.20 J (d) Chemical to heat transformation occurs in the operation of the circuit