Question

One point of the circuit is grounded (V = 0). What are the (a) size and (b) direction (up or down) of the current through resistance 1, the (c) size and (d) direction (left or right) of the current through resistance 2, and the (e) size and (f) direction of the current through resistance 3? (g) What is the electric potential at point A?

Answer 1

Answer:

(a) i₁ = 0.03818 A = 38.18 mA

(b) downward

(c) i₂ = 0.01091 A = 10.91 mA

(d) rightward

(e) i₃ = 0.02727 A = 27.27 mA

(f) leftward

(g) Eₐ = 3.818 Volts

Question:

The complete question is stated below and the figure is provided in the attachment:

In Fig. 27-47, E 1 = 6.00 V, E 2 = 12.0 V, R1 = 100 Ω,

R2 = 200 Ω, and R3 = 300 Ω. One point of the circuit is grounded

(V = 0). What are the (a) size and (b) direction (up or down) of the

current through resistance 1, the (c) size and (d) direction

(left or right) of the current through resistance 2, and the

(e) size and (f) direction of the current through resistance 3?

(g) What is the electric potential at point A?

Explanation:

Applying Kirchoff's voltage law in the loops of botg E₁ and E₂, in the clockwise and anti clockwise direction:

E₁ - i₂R₂ - i₁R₁ = 0  

E₂ - i₃R₃ - i₁R₁ = 0  

If, we apply Kirchhoff's current law at junction A, we get:

i₁ = i₂ + i₃

Using these relations in loop equations, and re-arranging:

E₁ - i₂R₂ - (i₂ + i₃) R₁ = 0     ___________ eqn (1)

E₂ - i₃R₃ - (i₂ + i₃) R₁ = 0    ___________ eqn (2)

Eqn (1) implies:

6 - 200 i₂ - 100 i₂ - 100 i₃ = 0

i₂ = (6 - 100i₃)/300

Eqn (2) implies:

12 - 300 i₃ - 100 i₂ - 100 i₃ = 0

12 - 400 i₃ = 100 i₂

using value of i₃ from eqn (1)

12 - 400 i₃ = (1/3)(6 - 100 i₃)

36 - 1200 i₃ = 6 - 100 i₃

1100 i₃ = 30

i₃ = 0.02727 A

using this value in eqn of  i₂:

i₂ = [6 - 100(0.02727)]/300

i₂ = (6 - 2.727)/300

i₂ = 0.01091 A

Since:

i₁ = i₂ + i₃

i₁ = 0.01091 A + 0.02727 A

i₁ = 0.03818 A

(a)

i₁ = 0.03818 A = 38.18 mA

(b)

Since, the value of current is positive, thus it will have the direction that was assumed.

Therefore, its direction will downward

(c)

i₂ = 0.01091 A = 10.91 mA

(d)

Since, the value of current is positive, thus it will have the direction that was assumed.

Therefore, its direction will rightward

(e)

i₃ = 0.02727 A = 27.27 mA

(f)

Since, the value of current is positive, thus it will have the direction that was assumed.

Therefore, its direction will leftward

(g)

With respect to the grounded portion, the potential drop at the resistance 1 will be equal to the potential at A Eₐ.

Therefore,

Eₐ = i₁R₁

Eₐ = (0.03818 A)(100 Ω)

Eₐ = 3.818 Volts