Finding the current in a complex circuit?

Question

For the circuit shown below, find the current I1. (Take emf1 = 59.4 V, emf2 = 110.4 V, R1 = 19.2 Ω, and R2 = 5.50 Ω)

 

http://www.webassign.net/colfunphys1/20-p-026-alt.gif

Answer 1

Draw 3 counter-clockwise current loops for the circuit given.

I₁ starts at E_110.4 and passes through R_19.2---R₂---R₆.

I₂ starts at E_59.4 and passes through R₆---R_5.5---R₃.

I₃ starts at R₂ and passes through R₃---R_5.5---R₆.

Write the simultaneous equations as

110.4 = 19.2I₁ + 2(I₁ − I₃) + 6(I₁ − I₃)

59.4 = 6I₂ + 5.5(I₂ − I₃) +3(I₂ − I₃)

0 = 8(I₁ − I₃) + 8.5(I₂ − I₃)

Rewrite these equations as

27.2I₁ + 0I₂ − 8I₃ = 110.4

0I₁ + 14.5I₂ − 8.5I₃ = 59.4

8I₁ + 8.5I₂ − 16.5I₃ = 0

Write a principal matrix for the coefficients of I₁, I₂, & I₃:

27.2----0------(-8)

0-----14.5----(-8.5)

8------8.5---(-16.5)

Write a second matrix where the first column of the

principal matrix is replaced by 110.4---59.4---0:

110.4------0-----------(-8)

59.4-----14.5--------(-8.5)

0----------8.5--------(-16.5)

Calculate the determinant of the principal matrix:

27.2×14.5×(-16.5) − 27.2×8.5.×(-8.5) − 0×0×(-16.5) + 0×8.5×(-8) + 8×0×(-8.5) − 8×14.5×(-8)

comes to -3614.4.

Calculate the determinant of the second matrix:

110.4×14.5×(-16.5) − 110.4×8.5.×(-8.5) − 59.4×0×(-16.5) + 59.4×8.5×(-8) + 0×0×(-8.5) − 0×14.5×(-8)

comes to -22476.

Then I₁ is obtained by (-22476/-3614.4) equal to 6.218459495 Ampères.