Question

Please help me figure out question 15 and explain it so i can understand it? look at the picture below for the question and information. thank you.

Answer 1

`17 Omega`

Explanation:

Since this is a series circuit, current in Amps must stay constant throughout the circuit.

So to find the Voltage at the `23Omega` resistor:
`V=IR`
`V=(3A)(23Omega)= 69 V`

Subtract the given voltage at that resistor from the total voltage to find the voltage at the resistor we don't know:
`120V-69 V= 51 V`

Now to find the resistance at the unknown resistor, remember the current is still constant:
`R= V/I`
`R= "51 V"/"3 A"= 17 Omega`

Answer 2

`R_"unknown" = 17 Omega`

Explanation:

I agree with Hi that the unknown resistor has resistance of 17 `Omega`. I will offer a different explanation. Go to this website:
https://www.maplesoft.com/content/EngineeringFundamentals/15/MapleDocument_15/Nodal%20and%20Loop%20Analysis.pdf
Scroll down to the section with the heading "Kirchhoff's Voltage Law (KVL)".

Kirchhoff's Voltage Law sums the voltage drops and gains across the 3 components. The discussion of the analysis of the circuit in Figure 2 applies well to your circuit. Read thru their discussion.

As you analyse the circuit in the clockwise direction, I encourage you to think of the voltage across the components. The voltage change as you compare the negative terminal to the positive terminal of the battery is +120 V. When the analysis takes you to a resistor, I encourage you to think not of the "voltage at the resistor" but of the "voltage drop across the resistor".

Ohm's Law allows us to determine the voltage across the `23Omega` resistor as `V = I*R = 3 A*23 Omega = 69 V`. When using Kirchhoff's Voltage Law, we need to determine whether the 69 V should have a + or a - sign. Since Kirchhoff says that the sum after analyzing a loop will be zero, and the battery has started us off with +120 V, the voltage change across resistors needs to be negative.

Going clockwise, notice that as the analysis encounters a resistor, the first terminal encountered is closer to the + terminal of the battery than is the other terminal. Therefore we have found a voltage drop with the value -69 V. Fortunate that it is negative since I said at the end of the previous paragraph that it "needs" to be negative.

Therefore the voltage loop equation is

`120 V - 69 V -3 A*R_"unknown" = 0`

Solving that:
`3 A*R_"unknown" = 120 V - 69 V`

`3 A*R_"unknown" = 51 V`

`R_"unknown" = (51 V)/ (3 A) = 17 Omega`

Disclaimer: This is an introductory example for Kirchhoff's Voltage Law. Application to a more complicated circuit will require more complication in the assignment of polarity the expression for the voltage across resistors.

I hope this helps,
Steve