This offline problem will walk you through an example of using the circuit equat

Question

This offline problem will walk you through an example of using the circuit equations. A real battery consists of an ideal battery and an internal resistor. Consider the following DC circuit which contains two real batteries connected in parallel to each other and then connected to an external resistor of resistance R. Battery 1, on the left, consists of an ideal battery of voltage epsilon and has internal resistance r. Battery 2, in the middle, consists of an ideal battery of voltage 2 epsilon and also has internal resistance r. Currents I_1, I_2, and I_3 have been drawn in and default current directions added. Note that if you wind up with a negative value for one of your currents, that just means that the current is flowing in the opposite direction. Choose one of the two junctions and write down the Junction Rule for that junction. Be sure to indicate which of the two junctions you are using! Choose two of the three closed loops and write down the Loop Rules for those loops. Be sure to indicate which two loops you are using! Be careful about signs! Let epsilon = 1.5 V, r = 1 ohm, and R = 100 ohm Solve for the three currents I_1, I_2, and I_3 (in amps). The terminal voltage of a battery is the voltage difference between the two terminals of a real battery (one that contains an internal resistance). That is, the terminal voltage takes into account both the voltage of the ideal battery and the voltage of the internal resistor. Find the terminal voltage of each of the batteries. How docs these terminal voltages compare? Why? Determine the power delivered by each of the two ideal batteries. Determine the power dissipated by each of the three resistors. Show that the total power delivered by the two ideal batteries equals the total power dissipated by the three resistors (conservation of energy!). Find the net power delivered by each of the real batteries.

Explanation / Answer

part a:

take the upper junction.

using kirchoff's current law,

I1+I2=I3

==>I1+I2-I3=0...(1)

part b:

loop rule in left loop:

E-I1*r+I2*r-2*E=0

==>(I1-I2)*r=-E...(2)

loop rule in right loop:

2*E-r*I2-R*I3=0

==>r*I2+R*I3=2*E...(3)

part c:

solving equations 1 ,2 and 3 using the values of E,r and R

we get

I1=-0.7438 A

I2=0.7562 A

I3=0.01243 A

part d:

terminal voltage of battery 1=E-I1*r

=1.5-1*(-0.7438)=2.2438 volts

terminal voltage of battery 2=2*E-r*I2

=2*1.5-1*0.7562=2.2438 volts

part e:

power delivered by battery 1=1.5*I1

=1.5*(-0.7438)=-1.1157 W

power delivered by battery 2=1.5*0.7562=1.14 W

part f:

power dissipated in r of battery 1=I1^2*r=0.55324 W

power dissipated in r of battery 2=0.7562^2*1=0.57183 W

power dissipated in R=I3^2*R=0.01545 W

part g:

total power by batteries=power delivered by battery 2=1.14 W

power consumed=0.55324+0.57183+0.01545=1.14052 W

so both the values are equal.

part h:

net power delivered by battery 1 =terminal voltage*I1

=2.2438*(-0.7438)=-1.66893 W

net power delivered by battery 2=2.2438*0.7562=1.69676 W

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