Solving Circuit Analysis Problem: Find V - Power of 5Ω Resistor is 20W

• Engineering
• hackett5
In summary, the problem involved finding the voltage across a 5Ω resistor with a power absorption of 20W. Using the equations V= I x R and P = V2/R, the voltage was found to be 10V. To find the total voltage supplied, the current going in and out of the node above the resistor was calculated using I= V/R. The total voltage was then found by adding the voltage across the 2Ω resistor to the 10V already known.

Homework Statement

Hello, I was hoping someone could walk me through the steps to solve this problem:

The power absorbed by the 5Ω resistor is 20W. Find V
http://img191.imageshack.us/img191/6411/14966985.jpg [Broken]

V= I x R
P = V2/R

The Attempt at a Solution

I first found the voltage across the 5Ω resistor to be 10V using V= I x R and substituting 20W / V for the current. How do I find the total voltage supplied?

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You know the voltage across the resistor, good. So you know the voltage of the node above the 5Ohm resistor. Can you tell us how much current goes in and out of that node? Where does it come from? Where does it flow to?

Well, using I= V/R, I get 2A (I1) going through the 5Ω node. The 2Ω and 3Ω resistors are in series so I can combine them as 5Ω. So through that node the current would also be 2A (I2) and voltage would be 10V. Then I2 + I1= Itotal or 2A +2A= 4A? The total resistance would be 2.5Ω and voltage = 10V?

So the total voltage would be: V= 10V + the voltage across the 2Ω resistor.

V= 10V + (2Ω x 4A)
V=18V

Bingo!

Awesome! Thanks for the help

1. How do I find the voltage across the 5Ω resistor?

The voltage across a resistor can be found using Ohm's Law, which states that V = IR (voltage = current x resistance). In this case, we know that the power (P) of the resistor is 20W, so we can use the formula P = VI (power = voltage x current) to solve for V. Rearranging the formula, we get V = P/I. Therefore, to find the voltage across the 5Ω resistor, we need to divide the power (20W) by the current (I) flowing through the resistor.

2. How do I calculate the current flowing through the 5Ω resistor?

To calculate the current through the resistor, we can use the formula I = P/V (current = power / voltage). We know that the power of the resistor is 20W and we can find the voltage across it using Ohm's Law. Once we have the voltage, we can plug it into the formula to solve for the current. Alternatively, we can also use the formula I = V/R (current = voltage / resistance), if we already know the voltage and resistance of the resistor.

3. Can I use Kirchhoff's Laws to solve this circuit analysis problem?

Yes, Kirchhoff's Laws can be used to solve this problem. Kirchhoff's Voltage Law (KVL) states that the sum of the voltages in a closed loop in a circuit must equal 0. Kirchhoff's Current Law (KCL) states that the sum of the currents entering and leaving a node (or junction) in a circuit must also equal 0. By applying these laws to the circuit, we can solve for the voltage and current of the 5Ω resistor.

4. What is the power dissipated by the 5Ω resistor?

The power dissipated by a resistor can be calculated using the formula P = VI (power = voltage x current). In this problem, we are given the voltage and we can calculate the current using Ohm's Law or Kirchhoff's Laws. Once we have both values, we can plug them into the formula to find the power dissipated by the resistor. In this case, the power is given as 20W, so we can also use the formula P = I^2R (power = current^2 x resistance) to solve for the current and then use the first formula to solve for the voltage.

5. How can I check my answer for this circuit analysis problem?

One way to check your answer is to use a circuit simulator program or an online circuit analysis tool. These tools allow you to input the circuit components and values and then calculate the voltage and current at each point in the circuit. You can compare your calculated values with the simulated values to see if they match. Another way to check your answer is to solve the problem using different methods (such as using different equations or laws) and see if you get the same result. If you do, then you can be more confident in your answer.