Question

5.00 a is flowing through a 10.0 ω resistor. how much power is being dissipated? 50.0w 250w 500w 2.5w question 4 (1 point) two 4.0-ω resistors are connected in parallel, and this combination is connected in series with 3.0 ω. what is the effective resistance of this combination? 1.2 ω 5.0 ω 7.0 ω 11 ω

Explanation:

First Question (Power Dissipation)

Step1: Recall the power formula for a resistor

The formula for power \( P \) dissipated in a resistor is \( P = I^2R \), where \( I \) is the current and \( R \) is the resistance.

Step2: Substitute the given values

Given \( I = 5.00 \, \text{A} \) and \( R = 10.0 \, \Omega \). Substitute into the formula: \( P=(5.00)^2\times10.0 \).

Step3: Calculate the power

First, \( (5.00)^2 = 25.0 \), then \( 25.0\times10.0 = 250 \, \text{W} \).

Step1: Calculate parallel resistance

For two resistors \( R_1 \) and \( R_2 \) in parallel, the formula is \( \frac{1}{R_{\text{parallel}}}=\frac{1}{R_1}+\frac{1}{R_2} \). Here, \( R_1 = R_2 = 4.0 \, \Omega \), so \( \frac{1}{R_{\text{parallel}}}=\frac{1}{4.0}+\frac{1}{4.0}=\frac{2}{4.0}=\frac{1}{2.0} \), thus \( R_{\text{parallel}} = 2.0 \, \Omega \).

Step2: Calculate series resistance

Now, this parallel combination is in series with \( R_3 = 3.0 \, \Omega \). For series resistors, \( R_{\text{total}} = R_{\text{parallel}}+R_3 \). Substitute \( R_{\text{parallel}} = 2.0 \, \Omega \) and \( R_3 = 3.0 \, \Omega \), so \( R_{\text{total}} = 2.0 + 3.0 = 5.0 \, \Omega \).

Answer:

250W

Second Question (Effective Resistance)