3. Ohm’s law states that the current flowing in a circuit varies directly as the applied voltage and inversely as the resistance. In an experiment, the resistance was increased by triple the percentage by which the voltage was increased. A student calculated the percent change in current to be 15%. Is the student correct?
• If yes, determine the percent change in voltage and resistance to three significant digits.
• If no, explain why and give an example of non-zero percent changes in V and R that would result in a 15% increase in I.
Maths - Henry, Monday, November 19, 2012 at 8:26pm
I1 = V/R.
Let V increase by 10%; then increase R
by 3 times as much(30%).
I2 = (V+0.1V)/(R+0.3R).
I2 = 1.1V/1.3R = 0.85V/R.
I2/I1 = (0.85V/R) / (V/R) = 0.85 = 85%.
%Change = 85%-100% = -15% = 15% Decrease
However, this value is valid for a 10%
increase in V only.
If V is increased by 20% and R increased by 3 times as much(60%),the
results is completely different:
I = 1.2V/1.6R = 0.75V/R. This is the
same as decreasing V by a factor of 0.75
while holding R constant. In the first
example, V was decreased by a factor of
0.85. Therefore, we cannot consider the student's calculation as being correct
unless the % increase in V was given.
I1 = V/R.
I2 = (V+3p*V)/(R+p*R).
I2 = V(1+3p)/R(1+p).
I2/I1 = (V(1+3p)/R(1+p))/(V/R).
The Vs and Rs cancel:
I2/I1 = (1+3p)/(1+p) = 1.15
(1+3p)/(1+p) = 1.15
Multiply both sides by (1+p):
1+3p = 1.15+1.15p
1.85p = 0.15
P = 0.081 = 8.1 % Increase in R.
3p = 3*.081 = 0.243 = 24.3% Increase in V.
I2 = 1.243V/1.081R = 1.15V/R.
I2/I1 = (1.15V/R) / (V/R) = 1.15 = 115%
%Change=115 - !00% = 15% = 15% Increase.