Question

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

Answer 1

The I3 will be 158 A.

How to find the current through the circuit?

• 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:

https://brainly.com/question/86531

#SPJ 10