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Is the graph of the relationship between V and I an example of direct or indirect proportion? Justify your answer.

Mathematics

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2 parts
1.2

V = 220I\text{V = 220I}

1.3

V = 220I\text{V = 220I}

Step-by-step solution

4 steps

Here are the answers to the questions based on the previous calculations for the 220Ω220 \Omega resistor.

Part 1.1

Is the graph of the relationship between V and I an example of direct or indirect proportion? Justify your answer.

The relationship between V and I is an example of direct proportion.
Justification: According to Ohm's Law (V=IRV = IR), when the resistance (RR) is constant, the voltage (VV) is directly proportional to the current (II). As the voltage increases, the current increases by the same factor, and the ratio VI\frac{V}{I} remains constant (equal to the resistance). A graph of V versus I would be a straight line passing through the origin.

Part 1.2

V = 220I\text{V = 220I}

Use the first and last two points from table/graph to determine the gradient of the graph and write the equation in the form of y=mx+cy = mx + c.

  1. Identify the first and last data points.

    From the previous calculations, the first point is (I1,V1)=(0.006818 A,1.5 V)(I_1, V_1) = (0.006818 \text{ A}, 1.5 \text{ V}) and the last point is (I6,V6)=(0.040909 A,9 V)(I_6, V_6) = (0.040909 \text{ A}, 9 \text{ V}).
    We will use II as the x-variable and VV as the y-variable, so the equation will be in the form V=mI+cV = mI + c.

  2. Calculate the gradient (mm).

    The gradient is given by the formula m=ΔVΔIm = \frac{\Delta V}{\Delta I}:

    m=V6−V1I6−I1m = \frac{V_6 - V_1}{I_6 - I_1} m=9 V−1.5 V0.040909 A−0.006818 Am = \frac{9 \text{ V} - 1.5 \text{ V}}{0.040909 \text{ A} - 0.006818 \text{ A}} m=7.5 V0.034091 Am = \frac{7.5 \text{ V}}{0.034091 \text{ A}} m≈220.00VAm \approx 220.00 \frac{\text{V}}{\text{A}} m≈220Ωm \approx 220 \Omega

  3. Determine the y-intercept (cc).

    Using the equation V=mI+cV = mI + c and the first point (0.006818,1.5)(0.006818, 1.5):

    1.5=(220)(0.006818)+c1.5 = (220)(0.006818) + c 1.5=1.49996+c1.5 = 1.49996 + c c=1.5−1.49996c = 1.5 - 1.49996 c≈0.00004c \approx 0.00004

    Since cc is very close to zero, we can consider it to be 0.

  4. Write the equation in the form y=mx+cy = mx + c.

    Substituting the values of mm and cc into V=mI+cV = mI + c:

    V = 220I\boxed{\text{V = 220I}}

Part 1.3

V = 220I\text{V = 220I}

Write V in terms of I by using the equation above.

Using the equation derived in 1.2:

V = 220I\boxed{\text{V = 220I}}
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