(3%) Problem 28: A 0.021- ammeter is placed in series with a 8.5- resistor in a
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(3%) Problem 28: A 0.021- ammeter is placed in series with a 8.5- resistor in a circuit. 33% Part (a) Calculate the resistance, in ohms, of the combination Grade Summary Deductions Potential 0% 100% Submissions Attempts remaining: 7 (2% per attempt) detailed view sin tan() | cotan asinO acosO) acotano cosh0tanh) cotanhO Degrees Radians atan()-1 | sinh() 0 END VO DELCLEAR Submit Hint I give up Hints: 0%-deduction per hint. Hints remaining: Feedback: 0% deduction per feedback. 33% Part (b) If the voltage is kept the same across the combination as it was through the 8.5- resistor alone, what is the percent decrease in current? 33% Part (c) increase in voltage? If the current is kept the same through the combination as it was through the 8.5- resistor alone, what is the percentExplanation / Answer
Total resistance in series=sum of individual resistances
So we have
Rnet=r1 +r2=0.021+8.5=8.521 ohm
B) :when only resistor was connected the current i was
I=v/R=v/8.5
Now when combined
I'=v/8.521
%decrease in current =delta i/i original *100
Where delta i is change in current
%decrease =(v/8.5 - v/8.521)/(v/8.5)*100
%decrease=0.246%
3);initially we have
V1=ir=i*8.5
When in combination =v2=i(8.521)
%increases =(v2-v1)/v2 *100
%increase={i(r1+r2) - (ir1)} /i(r1+r2) * 100
On simplifying the expression we get
% increase =r2/(r1+r2) *100
=0.021/(0.021+8.5)*100
=0.246%
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