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

In this circuit, the direction and size of currents through the resistances can be determined based on the voltage differences. The electric potential at point A is zero.

Step-by-step explanation:

In the given circuit, since one point is grounded (V = 0), we can determine the size and direction of the current through each resistance.

(a) The size of the current through resistance 1 is determined by the voltage difference across it and its resistance value using Ohm's Law (V = IR). Without the actual values, we cannot determine the size, only the direction. If the voltage source is connected to the left of resistance 1, the current will flow from left to right.

(b) The direction of the current through resistance 2 will depend on the voltage difference across it. If the positive terminal of the voltage source is connected to the top of resistance 2, the current will flow from top to bottom.

(c) Similarly, the direction of the current through resistance 3 will depend on the voltage difference across it. If the positive terminal of the voltage source is connected to the right side of resistance 3, the current will flow from right to left.

(g) The electric potential at point A is the same as the potential at the ground (V = 0).

Answer 2

(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?

Step-by-step 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