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Annette [7]
2 years ago
10

A circuit is constructed with six resistors and two batteries as shown. The battery voltages are V1 = 18 V and V2 = 12 V. The po

sitive terminals are indicated with a + sign, The values for the resistors are: R1 = R5 = 70 Ω, R2 = R6 = 106 Ω R3 = 59 Ω, and R4 = 83 Ω. The positive directions for the currents I1, I2 and I3 are indicated by the directions of the arrows. What is I3?

Physics
1 answer:
kaheart [24]2 years ago
7 0

The I3 will be 158 A.

<h3>How to find the current through the circuit?</h3>
  • The foundation of circuit analysis is Kirchhoff's circuit laws.
  • We have the fundamental instrument to begin studying circuits with the use of these principles and the equation for each individual component (resistor, capacitor, and inductor).
  • These rules aid in calculating the current flow in various network streams as well as the electrical resistance of a complicated network, or impedance in the case of AC.

To calculate I3 firstly, V4 has to be calculated,

V_{4} =I_{4} R_{4}

V_{4} = V_{2} / R_{4} + R_{5}  * R_{4}

V_{4} = 12 * 135 / 135+61

V_{4} = 8.26V

For I3,

I_{3} = R_{1} /(R1+R3 + (R1+R3)(R2+R6) * (V2 - V1 (R1+R2+R6/R1)

I3=(61)/((61)(50)+(61+50)(141+141)) (12 -18 (1+(141+141)/61)) = -.158 A

Hence, the current through I3 will be 158 A.

To learn more about Kirchoff's laws refer to:

brainly.com/question/86531

#SPJ 10

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