Problem 6. A d.c. circuit comprises three closed loops. Applying Kirchhoff's laws to the closed loops gives the following equations for current flow in milliamperes: 211 +312 – 413= 26 I1 – 512 – 313 = -87 -711 +212 +613= 12 Use determinants to solve for I1, I2 and I3.

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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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Problem 6. A d.c. circuit comprises three closed
loops. Applying Kirchhoff's laws to the closed
loops gives the following equations for current flow
in milliamperes:
211 +312 –413= 26
I - 512 – 313 = -87
-711 +212 +613= 12
Use determinants to solve for I1, I2 and I3.
Transcribed Image Text:Problem 6. A d.c. circuit comprises three closed loops. Applying Kirchhoff's laws to the closed loops gives the following equations for current flow in milliamperes: 211 +312 –413= 26 I - 512 – 313 = -87 -711 +212 +613= 12 Use determinants to solve for I1, I2 and I3.
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