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.
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.
Check:
I2 =(V+0.243V)/(R+0.081R)
I2 = 1.243V/1.081R = 1.15V/R.
I2/I1 = (1.15V/R) / (V/R) = 1.15 = 115%
%Change=115 - !00% = 15% = 15% Increase.
;
To determine if the student's calculation is correct, we need to analyze the given information and apply Ohm's law.
Let's assume the initial values of current (I), voltage (V), and resistance (R) are denoted as I₁, V₁, and R₁, respectively.
According to Ohm's law, the relationship between current, voltage, and resistance can be expressed as:
I = V / R
If the resistance is tripled, it becomes 3R₁. If the voltage is increased by some percentage, we can denote it as ΔV (change in voltage) and express it as:
V₂ = V₁ + ΔV
The resistance is increased by triple the percentage by which the voltage was increased. We can calculate this increase as:
R₂ = R₁ + 3R₁ * (ΔV/V₁)
The student claims that there is a 15% increase in current (I). To find if the student is correct, we can compare the initial and final current values:
(I₂ - I₁) / I₁ * 100 = 15%
Let's substitute the relevant values into this equation:
(V₂ / R₂ - V₁ / R₁) / (V₁ / R₁) * 100 = 15%
Now, let's simplify this equation further:
(V₂R₁ - V₁R₂) / (V₁R₂) * 100 = 15%
[(V₁ + ΔV)R₁ - V₁(R₁ + 3R₁ * (ΔV/V₁))] / [V₁(R₁ + 3R₁ * (ΔV/V₁))] * 100 = 15%
Now, let's simplify the equation even more:
ΔV / V₁ * 100 - 3ΔV / (1 + 3ΔV/V₁) * 100 = 15%
To determine if the student is correct, we need to solve this equation for ΔV (change in voltage). However, solving this equation analytically may not be straightforward, so we can use numerical methods or approximation techniques to find the value of ΔV.
If we find a solution where the percent change in voltage satisfies the above equation, then the student's calculation is correct. If there is no solution, then the student's calculation is incorrect.
So, without knowing the specific values of V₁, R₁, and ΔV, it is not possible to determine the percent change in voltage and resistance to three significant digits.
To summarize, in order to determine if the student's calculation is correct, we need to solve the equation derived from Ohm's law and the given information. Without the specific values of voltage, resistance, and change in voltage, it is not possible to obtain the percent change in voltage and resistance to three significant digits.