(596) Problem 15: of the batterics Consider a circuit shown in the figure. Ignor

Question

(596) Problem 15: of the batterics Consider a circuit shown in the figure. Ignore the internal resistances Randomized Variables 22 V 6212 V 1 2 Otheexpertta.com 14% Part (a) Express the current 11 going through resistor R1 in terms of the currents 12 and 13 going through resistors R2 and R3 Use the direction of the currents as specified in the figure 14% Part (b) Write the equation of potential change in loop EBAF in terms of the circuit elements #14% Part (c) Write the equation of potential change in loop DCAF in terms of the circuit elements 14% Part (d) Solve the three equations to get 13 - 14% Part (e) Calculate the numerical value of 13 in A 14% Part (f) Calculate the numerical value of 12 in A 14% Part (g) Calculate the numerical value of 11 in A

Explanation / Answer

part(a)

I1 = I2 + I3 .................(1)

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part(b)

e1 - I1*R1 - I3*R3 = 0.................(2)

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part(c)

-e2 + I2*R2 - I3*R3 = 0 .................(3)

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part(d)

using 1 in 2

e1 - (I2 + I3)*R1 - I3*R3 = 0

e1 - I2*R1 - I3*(R1 + R3) = 0 .........(4)

( eq3 *R1) + ( eq4* R2)

-e2*R1 + I2*R1*R2 - I3*R1*R3 = 0

e1*R2 - I2*R1*R2 - I3*R2*(R1+R3) = 0
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e1*R2 - e2*R1 - I3*(R1*R3 + R1*R2 + R2*R3 ) = 0

I3 = ( e1*R2 - e2*R1)/(R1*R2 + R2*R3 + R1*R3 )

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part(e)

I3 = ((22*7)-(12*5))/((5*7) + (7*7) + (5*7) )

I3 = 0.79 A

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part(f)

using I3 in eq 3

-e2 + I2*R2 - I3*R3 = 0

-12 + (I2*7) - (0.79*7) = 0

I2 = 2.5 A

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part(g)

I1 = I2 + I3

I1 = 0.79 + 2.5 = 3.29 A

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