The circuit shown in the figure below is connected for 1.70 min. (Assume R1 = 9.

Question

The circuit shown in the figure below is connected for 1.70 min. (Assume R1 = 9.00 , R2 1.80 , and V = 11.0 V.) 5.00 3.00 1.00 4.00 V (a) Determine the current in each branch of the circuit branch magnitude (A) direction left branch middle branch right branch -Select- Select-_ Select- (b) Find the energy delivered by each battery 4.00 V battery 11.0 V battery (c) Find the energy delivered to each resistor. 9.00 resistor 5.00 resistor 1.00 resistor 3.00 resistor 1.80 resistor (d) Identify the type of energy storage transformation that occurs in the operation of the circuit. This answer has not been graded yet. (e) Find the total amount of energy transformed into internal energy in the resistors kJ

Explanation / Answer

Let

IL=Current flowing in left branch

IM=Current flowing in middle branch

IL=Current flowing in right branch

a)

From Kirchoff's Junction rule

IR-IL-IM=0----------------1

Applying kirchoff's loop rule on right

11-4.8IR-6IM-4=0

4.8IR+6IM=7-------------2

Applying kirchoff's loop rule on left

-6IM-4+9IL=0

9IL-6IM=4--------------3

solving 1 ,2 and 3 we get

IL=0.6762 A

IM=0.3476 A

IR=1.0238 A

b)

E4=VIMt =-4*0.3476*(1.7*60)=-141.8 J

E11=11*1.0238*(1.7*60)=1.148 KJ

c)

E9=IL2Rt=0.67622*9*(1.7*60)=419.8 J

E5=0.34762*5*(1.7*60)=61.6 J

E1=0.34762*1*(1.7*60)=12.3 J

E3=1.02382*3*(1.7*60)=320.7 J

E1.8=1.02382*1.8*(1.7*60)=192.4 J

d)

chemical energy in batteries is transformed into internal energy in resistors

e)

E=419.8+61.6+12.3+320.7+192.4

E=1.007 KJ

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