The circuit shown in the figure below is connected for 1.70 min. (Assume R_1 = 6

Question

The circuit shown in the figure below is connected for 1.70 min. (Assume R_1 = 6.80 Ohm, R_2 = 1.20 Ohm, and V = 11.0 V.) Determine the current in each branch of the circuit. Find the energy delivered by each battery. 4.00 V battery J 11.0 V battery kJ Find the energy delivered to each resistor. 6.80 Ohm resistor J 5.00 Ohm resistor J 1.00 Ohm resistor J 3.00 Ohm resistor J 1.20 Ohm resistor J Identify the type of energy storage transformation that occurs in the operation of the circuit. Chemical energy in the 17.0-V battery is transformed into internal energy in the resistors. The 4.00-V battery is being charged, so its chemical potential energy is increasing at the expense of some of the chemical potential energy in the 17.0-V battery. Find the total amount of energy transformed into internal energy in the resistors.

Explanation / Answer

Let currents through left , middle and right branch be I1 , I2 and I3 , in upward direction respectively.
By Krichoff's law junction rule
I1 + I2 +I3 = 0 ..........1
Applying loop rule in the left loop, in anticlock wise direction we get
4 - 6 (I2) + 6.8(I1) = 0 ..........2
Applying loop rule in the right loop, in anticlock wise direction we get
11 -4.2 (I3) + 6(I2) -4 = 0 ........3
Solving these 3 equations we get
a) I1 = -0.88, I2 = - 0.32 and I3 = 1.20
   -ve sign indicates current is in downward  direction in left and middle branch.

b) Energy delivered by battery is
   VIt , if current inside it is from -ve terminal to + ve terminal and
   - VIt  if current inside it is from +ve terminal to -ve terminal or energy is consumed by battery.
Energy delivered by 11 V battery = 11x 1.2 x (1.7 x 60) = 1.35 KJ
Energy delivered by 4 V battery = -4 x 0.32 x (1.7 x 60) = 130.6 J

c) Energy delivered ti resistor is I2Rt. Hence
Energy delivered to 6.8 ohm = 0.882x6.8x1.7x60 = 537.1 J
   5.0 ohm = 0.322x5x1.7x60 = 52.2 J
   1.0 ohm = 10.4 J
   3.0 ohm = 1.22x3x1.7x60 = 440.6 J
1.2 ohm = 176.3 J

d) 1.35 KJ of chemical energy of 11 V battery is converted to 130.6 J of chemical energy of 4V battery and remaining into internal energy of resistors.

e) It is sum of energy delivered to all resistors that is 1.2 KJ

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