In the figure the current in resistance 6 is i6 = 1.48 A and the resistances are R1 = R2 = R3 = 1.85 Ω, R4 = 15.3 Ω, R5 = 8.11 Ω, and R6 = 3.88 Ω. What is the emf of the ideal battery?

Question

In the figure the current in resistance 6 is i₆ = 1.48 A and the resistances are R₁ = R₂ = R₃ = 1.85 Ω, R₄ = 15.3 Ω, R₅ = 8.11 Ω, and R₆ = 3.88 Ω. What is the emf of the ideal battery? 3 significant digits.

Link for the figure: https://imgur.com/IqBSWUN

Answer 1

Reg. I'd be happy to help you work this problem.

Basically, the fundamental principles that are involved here are:

1) Conservation of electrical charge; and

2) Conservation of energy.

In the case of electric circuits with lumped parameter components, these two fundamental principles are Kirchhoff's laws.

Kirchhoff's current law - the current going into a node (a junction) equals the current coming out of a node.

Kirchhoff's voltage law - the total voltage rise and voltage drop in any closed loop in a circuit is zero. The voltage change through a resistor is negative (a drop) if we move through a resistor in the direction of the current. The voltage change through a resistor is positive (a rise) if we move through a resistor opposite the direction of the current.

Finally, the directions (algebraic signs) of currents can be assumed for convenience when setting up a problem. If the solution give us a positive value for an unknown current then the current direction is as we drew in the diagram. If the solution gives is a negative value for an unknown current then the current direction is opposite of what we drew in the diagram.

Let

i₁ be the current in R₁

i₂ be the current in R₂

i₃ be the current in R₃

i₄ be the current in R4

i₅ be the current in R₅

i₆ be the current in R₆

E be the EMF or the voltage in the battery.

We are given:

R₁=R₂=R₃=1.85Ω

R₄=15.3Ω

R₅=8.11Ω

R₆=3.88Ω

i₆=1.48A

The unknown values in the problem are i₁, i₂, i₃, i₄, i₅, and E. We have 6 unknowns so we need 6 independent linear equations to solve the problem.

Applying Kirchhoff's current law to all the nodes in the diagram we have:

1) i₁=i₂+i₃

2) i₂=i₄+i₅

3) i₅=i₆=1.48A

3) implies

2) i₂=i₄+i₆

Equation 3 solves for i₅ so we have 5 unknowns and we need 5 independent linear equations to solve.

Define ground (where we assume the voltage is zero) to be the negative terminal of the battery E.

In the diagram, let:

Loop 1 be the loop that goes from ground clockwise through E, through R₁, and through R₃ back to ground.

Loop 2 be the loop that goes from ground clockwise through E, through R₁, through R₂, and through R₄ back to ground.

Loop 3 be the loop that goes from ground clockwise through E, through R₁, through R₂, through R₅, and R₆ back to ground.

Loop 4 be the loop that goes from ground clockwise through R₃, through R₂, and through R₄ back to ground.

Loop 5 be the loop that goes from ground clockwise through R₄, through R₅, and through R6 back to ground.

Loop 6 be the loop that goes from ground clockwise through R₃, through R₅, and through R6 back to ground.

From this we have a total of 8 equations:

Eq 1 i₁=i₂+i₃

Eq 2 i₂=i₄+i₆

Eq 3 E-i₁R₁-i₃R₃=0

Eq 4 E-i₁R₁-i₂R₂-i₄R₄=0

Eq 5 E-i₁R₁-i₂R₂-i₆R₅-i₆R₆=0

Eq 6 i₃R₃-i₂R₂-i₄R₄=0

Eq 7 i₄R₄-i₆R₅-i₆R₆=0

Eq 8 i₃R₃-i₂R₂-i₆R₅-i₆R₆=0

You could us a matrix calculator but simple math is enough. If we examine the 8 equations in light of the fact that i₆ is known (1.48A) we see the following chain of simple relations:

Eq 4 and 5 imply i₄=i₆(R₅+R6)/R₄=1.16A

Eq 2 then implies i₂=i₄+i₆=2.64A

Eq 6 then implies i₃=(i₂R₂+i₄R₄)/R₃=12.23A

Eq 1 then implies i₁=i₂+i₃=14.87A

And finally,

Eq 2 then implies E=i₁R₁+i₃R₃=50.14V

Note that since all of the current values we solved for were algebraically positive, we got the directions of our currents correct.

I hope this helped!