The total current I = I 1 + I 2 in a circuit is measured as I 1 = 150 ± 1A, I 2 = 250 ± 2A, where the limits of error are given as standard deviations, and Current I is measured as -
(400 ± 2.24) A
Current Measurement in Circuits: Error Propagation
In electrical circuits, when combining measurements, it is essential to understand how uncertainties or errors from individual measurements affect the overall result. This problem asks us to determine the total current \(I\) in a circuit, which is the sum of two individual currents, \(I_1\) and \(I_2\). The errors for both \(I_1\) and \(I_2\) are provided as standard deviations.
The relationship between the total current and the individual currents is given as: \[ I = I_1 + I_2 \] We are given the following measured values with their associated standard deviations:
- Current \(I_1 = 150 \pm 1A\)
- Current \(I_2 = 250 \pm 2A\)
Based on the problem statement, the limits of error are explicitly given as standard deviations. Therefore:
- Standard deviation of \(I_1\), denoted as \(\sigma_1 = 1A\)
- Standard deviation of \(I_2\), denoted as \(\sigma_2 = 2A\)
Total Current Calculation (Nominal Value)
The first step is to calculate the nominal, or most probable, value of the total current \(I\). This is simply the sum of the nominal values of \(I_1\) and \(I_2\):
\[ I_{\text{nominal}} = I_1^{\text{nominal}} + I_2^{\text{nominal}} \] \[ I_{\text{nominal}} = 150A + 250A \] \[ I_{\text{nominal}} = 400A \]Error Propagation: Standard Deviation of Total Current
When two or more independent quantities are added or subtracted, and their uncertainties are expressed as standard deviations, the standard deviation of the resulting sum or difference is calculated using the "root sum of squares" method. This method is also known as adding errors in quadrature.
The formula for the standard deviation of the total current (\(\sigma_I\)) when \(I = I_1 + I_2\) is:
\[ \sigma_I = \sqrt{\sigma_1^2 + \sigma_2^2} \]Now, let's substitute the given standard deviations (\(\sigma_1 = 1A\) and \(\sigma_2 = 2A\)) into this formula:
\[ \sigma_I = \sqrt{(1A)^2 + (2A)^2} \] \[ \sigma_I = \sqrt{1^2 + 2^2} \] \[ \sigma_I = \sqrt{1 + 4} \] \[ \sigma_I = \sqrt{5} \]To get the numerical value, we calculate the square root of 5:
\[ \sigma_I \approx 2.236067... A \]Rounding this value to two decimal places, which is a common practice for measurement uncertainties and matches the precision in the options provided:
\[ \sigma_I \approx 2.24 A \]Final Total Current Measurement Result
Combining the nominal total current value and its calculated standard deviation, the total current \(I\) is measured as:
\[ I = (I_{\text{nominal}} \pm \sigma_I) A \] \[ I = (400 \pm 2.24) A \]This result indicates that the total current in the circuit is approximately 400 A, with a standard deviation (uncertainty) of about 2.24 A.
Important Questions from Miscellaneous
A stone is thrown horizontally from the top of a 20 m high building with a speed of 12 m/s. It hits the ground at a distance R from the building. Taking g = 10 m/s2 and neglecting air resistance will give :
A sphere of volume V is made of a material with lower density than water. While on Earth, it floats on water with its volume f1V (f1 < 1) submerged. On the other hand, on a spaceship accelerating with acceleration a < g (g is the acceleration due to gravity on Earth) in outer space, its submerged volume in water is f2V. Then:
A railway wagon (open at the top) of mass M1 is moving with speed v1 along a straight track. As a result of rain, after some time it gets partially filled with water so that the mass of the wagon becomes M2 and speed becomes v2. Taking the rain to be falling vertically and the water stationery inside the wagon, the relation between the two speeds v1 and v2 is :
Consider the following statements:
1. Distance between the longitudes becomes zero on North Pole and South Pole.
2. Distance between the longitudes is maximum on the Equator.
3. Number of longitudes is more than number of latitudes.
Which of the statements given above is/are correct?
One block of 2⋅0 kg mass is placed on top of another block of 3⋅0 kg mass. The coefficient of static friction between the two blocks is 0⋅2. The bottom block is pulled with a horizontal force F such that both the blocks move together without slipping. Taking acceleration due to gravity as 10 m/s2, the maximum value of the frictional force is :
