Question

Solve the current and voltage problems for the ci

Answer 1

Explanation:

Each resistor has a resistance of R.

In the first problem, each row of three resistors in series has a resistance or 3R.

A₂ = 4A, so the voltage drop across the rows is:

V = IR

V = (4)(3R)

V = 12R

The voltage drop equals the voltage gain (Kirchoff's voltage law).  The battery has a voltage of 9V, so:

9 = 12R

R = 0.75 Ω

The three rows are in parallel with each other, so the total resistance is:

∑R = (1/(3R) + 1/(3R) + 1/(3R))⁻¹

∑R = R

∑R = 0.75 Ω

The current A₁ can be found with Ohm's law:

V = IR

9 = (A₁) (0.75)

A₁ = 12 A

Using Ohm's law to find V₁, V₂, and V₃:

V = IR

V₁ = (4)(0.75)

V₁ = 3V

V₂ = (4)(2×0.75)

V₂ = 6 V

V₃ = (4)(3×0.75)

V₃ = 9 V

In the second problem, all the resistors are in series.  So the total resistance is:

∑R = 8R

The battery voltage is 16V, and A₂ = 6A.  So using Ohm's law:

V = IR

16 = (6)(8R)

R = 1/3 Ω

Since there's only one loop, the current is the same at all points.  So A₁ = A₂ = 6A.

Using Ohm's law to find each voltage:

V = IR

V₁ = (6)(1/3)

V₁ = 2V

V₂ = (6)(2×1/3)

V₂ = 4 V

V₃ = (6)(4×1/3)

V₃ = 8 V

V₄ = (6)(2×1/3)

V₄ = 4 V

Answer 2

For the circuit shown in the left side of the image, A₁=12A, V₁=3V, V₂=6V, and V₃=9V.

The voltage source is Vs = 3(3V) = 9V is connected in parallel to the three branches that each contain three equals light bulbs, the three branches are in parallel meaning that the voltage in each branch is 9V.

The current through one branch is A₂ = 4A, we can calculate the equivalent resistance in that branch using Ohm's Law:  

Req = V/I, where I = A₂ = 4A

Req = 9V/4A = 2.25Ω

We can calculate the resistance in each light bulb, there are three light bulb in each branch that have a equivalent resistance of 2.25Ω, the light bulbs are in serie, so:

Req = R + R + R = 3R

The resistance of each light bulb is R = Req/3 = 2.25Ω/3 = 0.75Ω

To calculate V₁, using Ohm's Law:

V₁ = I R = (4A)(0.75Ω) = 3V

Then, the other branches are in parallel and it has the same amount of light bulbs and the same characteristics, each branch has a equivalent resistance of 2.25Ω, and each light bulb has a resistance of 0.75Ω.

To calculate V₂ and V₃:

We know that the voltage in each branch is 9V, each light bulb has a drop of voltage of 3V. So, the total voltage in one branch is the sum of each drop of voltage on each light bulb Vt = 3V + 3V + 3V = 9V.

V₂ = 3V + 3V = 6v

v₃ = Vt = 3V + 3V + 3V = 9V

To calculate the total current A₁ through the circuit, we know that the current through each branch is 4A, the total current A₁ is:

A₁ = 3A₂ = 3(4A) = 12A

For the circuit shown in the right side of the image, A₁=A₂=6A, V₁=2V, V₂=V₄=4V, and V₃=8V.

The voltage source is Vs = 4(-4V) = -16V connected in series with 8 light bulbs, the current flow through each light bulb, It= A₁ = A₂ = 6A. The voltage source is Vs= -16V, the '-' means that the current flows in the opposite direction.

To calculate the equivalent resistance of the circuit, using Ohm's Law:

Req = Vs/It = 16V/6A = 2.67Ω

The resistance of each light bulb is R = Req/8 = 2.67Ω/8 = 0.33Ω

The drop of voltage on each light bulb is:

V = It R = (6A)(0.33Ω) = 2V

V₁ = 2V

V₂ = V₄ = 2(2V) = 4V

V₃ = 4(2V) = 8V