Question

Consider the circuit shown below. (Due to the nature of this problem, do not use rounded intermediate values in your calculations-including answers submitted in WebAssign.) 1₁ 12 13 14 15 || = = R₁ = 70 (a) Find 1₁, 12, 13, 14, and 15 (all in A). (Indicate the direction with the signs of your answers.) A A A A A = V₁ = 13 V R₂ = 90 14₁ R3 = 60 (b) Find the power supplied by the voltage sources (in W). W R₁ = 60 V/₂=4V (c) Find the power dissipated by the resistors (in W). W 15

Answer 1

In this circuit, Ohm's Law can be used to find the unknown currents and calculate the power supplied by the voltage sources and the power dissipated by the resistors.

Step-by-step explanation:

In order to solve this problem, we need to use Ohm's Law, which states that the current (I) flowing through a circuit is equal to the voltage (V) divided by the resistance (R). In a series circuit, the current is the same through all components, while the voltage divides across each component. Let's solve for the unknown currents:

I1: Since the resistors R1, R2, and R3 are all in series, the total resistance (R) is equal to the sum of their individual resistances. Therefore, R = R1 + R2 + R3 = 70 + 90 + 60 = 220 ohms. Now, we can use Ohm's Law to find I1: I1 = V1/R = 13/220 = 0.059 A (rounded to three decimal places).

I2: Since R2 and R3 are in parallel, the total resistance (R) can be calculated using the formula 1/R = 1/R2 + 1/R3. Plugging in the values, we get: 1/R = 1/90 + 1/60 = 1/54. Solving for R gives us R = 54 ohms. Now, we can find I2 using Ohm's Law: I2 = V2/R = 4/54 = 0.074 A (rounded to three decimal places).

I3: Since R2 and R3 are in parallel, the total resistance (R) is still 54 ohms. Using Ohm's Law, we can find I3: I3 = V2/R = 4/54 = 0.074 A (rounded to three decimal places).

I4: Since R4 is in series with the parallel combination of R2 and R3, the total resistance is equal to R4 + R = 54 + 60 = 114 ohms. Using Ohm's Law, we can find I4: I4 = V2/R = 4/114 = 0.035 A (rounded to three decimal places).

I5: Since R5 is in parallel with the series combination of R2, R3, and R4, the total resistance is given by the formula 1/R = 1/R4 + 1/R5. Plugging in the values, we get: 1/R = 1/114 + 1/15. Solving for R gives us R = 12.576 ohms. Now, we can find I5 using Ohm's Law: I5 = V2/R = 4/12.576 = 0.318 A (rounded to three decimal places).

(b) To find the power supplied by the voltage sources, we can use the formula P = IV, where P is power, I is current, and V is voltage. So the power supplied by each voltage source is:

Power supplied by V1: P1 = I1 * V1 = 0.059 A * 13 V = 0.767 W (rounded to three decimal places).

Power supplied by V2: P2 = (I2 + I3 + I4 + I5) * V2 = (0.074 A + 0.074 A + 0.035 A + 0.318 A) * 4 V = 1.539 W (rounded to three decimal places).

(c) To find the power dissipated by the resistors, we can use the formula P = I^2R, where P is power, I is current, and R is resistance. So the power dissipated by each resistor is:

Power dissipated by R1: PR1 = I1^2 * R1 = (0.059 A)^2 * 70 Ω = 0.249 W (rounded to three decimal places).

Power dissipated by R2: PR2 = (I2 + I3 + I4 + I5)^2 * R2 = (0.074 A + 0.074 A + 0.035 A + 0.318 A)^2 * 90 Ω = 1.394 W (rounded to three decimal places).

Power dissipated by R3: PR3 = (I2 + I3 + I4 + I5)^2 * R3 = (0.074 A + 0.074 A + 0.035 A + 0.318 A)^2 * 60 Ω = 0.928 W (rounded to three decimal places).

Power dissipated by R4: PR4 = I4^2 * R4 = (0.035 A)^2 * 60 Ω = 0.073 W (rounded to three decimal places).

Power dissipated by R5: PR5 = I5^2 * R5 = (0.318 A)^2 * 15 Ω = 1.807 W (rounded to three decimal places).

Answer 2

In the given circuit, we are asked to find the currents (1₁, 12, 13, 14, and 15) in Amperes and the power supplied by the voltage sources and power dissipated by the resistors in Watts.

To solve for the currents in the circuit, we can use Ohm's Law and apply Kirchhoff's laws.

First, we can calculate the total resistance (R_total) of the parallel combination of resistors R₂, R₃, and R₁. Since resistors in parallel have the same voltage across them, we can use the formula:

1/R_total = 1/R₂ + 1/R₃ + 1/R₁

Once we have the total resistance, we can find the total current (I_total) supplied by the voltage sources by using Ohm's Law:

I_total = V₁ / R_total

Next, we can find the currents through the individual resistors by applying the current divider rule. The current through each resistor is determined by the ratio of its resistance to the total resistance:

I₁ = (R_total / R₁) * I_total

I₂ = (R_total / R₂) * I_total

I₃ = (R_total / R₃) * I_total

To calculate the power supplied by the voltage sources, we use the formula:

Power = Voltage * Current

Therefore, the power supplied by the voltage sources can be found by multiplying the voltage (V₁) by the total current (I_total).

Finally, to find the power dissipated by each resistor, we can use the formula:

Power = Current^2 * Resistance

Substituting the respective currents and resistances, we can calculate the power dissipated by each resistor.

By following these steps, we can find the currents (1₁, 12, 13, 14, and 15) in the circuit, as well as the power supplied by the voltage sources and the power dissipated by the resistors.

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