Class I Help Ch. 18 Homework, pt. 4 Begin Date: 9/15/2017 2:00:00 PM-Due Date: 9

Question

Class I Help Ch. 18 Homework, pt. 4 Begin Date: 9/15/2017 2:00:00 PM-Due Date: 9/18/2017 5:00:00 PM End Date: 9/20/2017 9-39:00 AM (17%) Problem 1: The current in a simple single resistor (of resistance R) circuit is/,-5.9 A. A second resistor, R2-68 .is added to the first in parallel and the current through the battery increases to 12-9.6 A nt Status ere for view > 33% Part (a) Input an expression for the resistance of R1. Grade Summary Deductions Potential 92% Status Partial Attempts remaining 4 4 5 6 1 2 3 detailed view 0 4% Submit Hint I give up! Hints:1% dedaction per hit. Hints remaining: 2 Feedback: 1deduction per feedback 33% Part (b) what is the resistance in m 33% Part (c) Can the current in a circuit ever decrease by adding a resistor in parallel?

Explanation / Answer

According to the question

We know that

potential difference across R1 in first case

=potential difference across R1 and R2 in series in second case

Then,

A)

=I1*R1=I2*(R1+R2)

=R1*(I1-I2)=I2*R2

=R1=I2*R2/(I1-I2)

Then

B)

using values of I2, I1 and R2 we get
R1=9.6*68/(5.9-9.6)=-163. 2ohms

Then,

C)

V=I2*(R1+R2)

=R1+R2=V/I2

=R2=(V/I2)-R1

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