Help with Self Test Question from Basic Electricity (Valkenburgh)

[troutnut68]

Hi Guys,

I'm a newbie working my way through Basic Electricity(1992) by Van Valkenburgh which is an old out of print book.  I've run into a Self Test problem that I cannot completely solve.  It is on page 2-87 and regards finding Voltage, Current and Resistance in a Parallel Circuit.  The question is part b and asks you to find Voltage E, current I1 and I2 and resistance R2.  I came up with E=300 V and I1= 1 Ampere but cannot figure out how to get R2 and I2.  The attachment is a camera picture of the problem.  Is there enough information to solve for R2 and I2 or is this test question broken?

E=R4 x I4 = 100 ohms x 3 A = 300 Volts     

I1 =  E / R1  =  300 V  /  300 ohms  =  1 A

R2 = ?

I2  =  ?

Thanks for any help offered,

Alan

[IanB]

Hi there,

In the absence of further information, it does not look like there is a way to find R2 and I2.

We know E as you have deduced, and we can write the equation E = R2 * I2. This is one equation with two unknowns. More information would be needed to pin down either of R2 or I2 to go any further. A common way for such problems to be set up is that the total current in the loop is given, but that doesn't seem to be the case here.

[Ian.M]

Also note there is *NO* R3 in the part (b) circuit, so it seems likely that the author accidentally submitted an incomplete circuit with the draft and the proof-reader(s) didn't notice.

[tpowell1830]

I don't see the text part of the question, so does the text mention the total current of the circuit?

[troutnut68]

Thanks for the responses to my question.  No, there was no further information for this question other than fill in the unknowns.  This is obviously a broken question that made it into the book and I will consider it case closed.  I'm moving on to the next part of the book.

Thanks again,

Alan

[Terry Bites]

Convert your reistances to condcutances (1/R =S) and then its a simple matter of addition and subtraction. When you have your mystery conductance worked out, just take the inverse (1/S).
I=E*S

Van V as in Common Core?

[troutnut68]

If total current was provided it would be simple to solve the rest of it.