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This is electricity and MAGNETISM practical questions. Solve all systematically. Create the exact graph, the way it is supposed to be plot.

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Answer

c

Approximately 3906Ω\text{Approximately } 3906 \Omega

Step-by-step solution

4 steps

This problem involves a meter bridge circuit. The total length of the meter bridge wire is l=1000 mml = 1000 \text{ mm}. The known resistor is R=680ΩR = 680 \Omega. The table provides values for the unknown resistor RxR_x and the corresponding balance length lxl_x.

A critical observation: In a meter bridge, as the unknown resistance RxR_x increases, the balance length lxl_x (measured from the end connected to RxR_x) should also increase. However, the provided data shows that as RxR_x increases, lxl_x decreases. This indicates an inconsistency in the experimental data or how it's presented.

Given the instruction to "Plot a graph of RxR_x against (l−lx)/lx(l-l_x)/l_x" and "Determine the slope of your graph and state its significance," it is implied that a linear relationship is expected. The standard meter bridge equation is RxR=lxl−lx\frac{R_x}{R} = \frac{l_x}{l-l_x}, which implies Rx=Rlxl−lxR_x = R \frac{l_x}{l-l_x}. If we plot RxR_x against l−lxlx\frac{l-l_x}{l_x}, this would result in a curve (Y=R⋅X−1Y = R \cdot X^{-1}).

To obtain a linear graph as typically expected in such problems, we must assume the meter bridge equation is interpreted as Rxl−lx=Rlx\frac{R_x}{l-l_x} = \frac{R}{l_x}, which rearranges to Rx=R(l−lxlx)R_x = R \left( \frac{l-l_x}{l_x} \right). Under this assumption, plotting RxR_x (Y-axis) against l−lxlx\frac{l-l_x}{l_x} (X-axis) would yield a straight line passing through the origin with a slope equal to RR. We will proceed with this assumption to fulfill the requirement of finding a slope.

Part a· 2 steps

Find (l−lx)/lx(l-l_x)/l_x. Tabulate your results.

  1. Calculate (l−lx)(l-l_x) for each lxl_x value, where l=1000 mml = 1000 \text{ mm}.

  2. Calculate (l−lx)/lx(l-l_x)/l_x for each pair.

    Convert RxR_x from kΩk\Omega to Ω\Omega.

    Rx/Ωlx/mml−lx/mm(l−lx)/lx10008401000−840=160160840≈0.19015007021000−702=298298702≈0.42522006181000−618=382382618≈0.61835005321000−532=468468532≈0.88045004561000−456=544544456≈1.193\begin{array}{|c|c|c|c|} \hline R_x/\Omega & l_x/\text{mm} & l-l_x/\text{mm} & (l-l_x)/l_x \\ \hline 1000 & 840 & 1000-840=160 & \frac{160}{840} \approx 0.190 \\ 1500 & 702 & 1000-702=298 & \frac{298}{702} \approx 0.425 \\ 2200 & 618 & 1000-618=382 & \frac{382}{618} \approx 0.618 \\ 3500 & 532 & 1000-532=468 & \frac{468}{532} \approx 0.880 \\ 4500 & 456 & 1000-456=544 & \frac{544}{456} \approx 1.193 \\ \hline \end{array}

Part b

Plot a graph of RxR_x against (l−lx)/lx(l-l_x)/l_x.

To plot the graph:

  • X-axis: (l−lx)/lx(l-l_x)/l_x (dimensionless). Scale: 0 to 1.4 (e.g., 10 small squares = 0.2 units).
  • Y-axis: Rx/ΩR_x/\Omega. Scale: 0 to 5000 Ω\Omega (e.g., 10 small squares = 500 Ω\Omega).
  • Plot the points:
    • (0.190,1000)(0.190, 1000)
    • (0.425,1500)(0.425, 1500)
    • (0.618,2200)(0.618, 2200)
    • (0.880,3500)(0.880, 3500)
    • (1.193,4500)(1.193, 4500)
  • Draw the best-fit straight line through these points.

(Note: Due to the inconsistency in the provided data, the points may not fall perfectly on a straight line. A best-fit line should be drawn to represent the general trend.)

Part c

Approximately 3906Ω\text{Approximately } 3906 \Omega

Determine the slope of your graph and state its significance.

  1. Choose two points on the best-fit line (not necessarily data points) that are far apart to calculate the slope.

    Let's pick two points from the calculated table that are roughly on the line.
    Using points (X1,Y1)=(0.425,1500)(X_1, Y_1) = (0.425, 1500) and (X2,Y2)=(1.193,4500)(X_2, Y_2) = (1.193, 4500):

    Slope=ΔYΔX=Y2−Y1X2−X1\text{Slope} = \frac{\Delta Y}{\Delta X} = \frac{Y_2 - Y_1}{X_2 - X_1}
  2. Calculate the slope.

    Slope=4500Ω−1500Ω1.193−0.425=3000Ω0.768≈3906.25Ω\text{Slope} = \frac{4500 \Omega - 1500 \Omega}{1.193 - 0.425} = \frac{3000 \Omega}{0.768} \approx 3906.25 \Omega

    The calculated slope is approximately 3906Ω3906 \Omega. This value is significantly different from the given known resistance R=680ΩR = 680 \Omega. This discrepancy arises from the inconsistency in the provided experimental data. However, following the assumption that Rx=R(l−lxlx)R_x = R \left( \frac{l-l_x}{l_x} \right), the slope of the graph represents the known resistance RR.

    • Slope: Approximately 3906Ω\boxed{\text{Approximately } 3906 \Omega}
    • Significance: The slope of the graph represents the value of the known resistor RR.

Part d

State two precautions you would have taken if you were to perform the experiment in a laboratory.

  • Ensure all connections are tight to minimize contact resistance.
  • Avoid overheating the wire by keeping the current on only when taking readings, to prevent changes in resistance.
  • Use a sensitive galvanometer to accurately detect the null point.
  • Repeat readings and take an average to reduce random errors.
  • Ensure the meter rule is parallel to the wire to avoid parallax error in length measurements.

Part e

Explain how an ammeter is connected in an electric circuit.

An ammeter is always connected in series with the component through which the current is to be measured. It has a very low internal resistance to minimize its effect on the circuit's total resistance and thus on the current being measured.

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