A d'Arsonal meter with an internal resistance of 1 kohm requires 10 mA to produce full-scale deflection. Calculate thew value of
1 answer:
<h2>Question:</h2>
A d’Arsonval meter with an internal resistance of 1 kΩ requires 10 mA to produce full-scale deflection. Calculate the value of a series resistance needed to measure 50 V of full scale.
<h2>Answer:</h2>4kΩ
<h2>Explanation:</h2>Given;
internal resistance, r = 1kΩ
current, I = 10mA = 0.01A
Voltage of full scale, V = 50V
Since there is full scale voltage of 50V, then the combined or total resistance (R) of the circuit is given as follows;
From Ohm's law
V = IR
R = [substitute the values of V and I]
R =
R = 5000Ω = 5kΩ
The combined resistance (R) is actually the total resistance of the series arrangement of the series resistance() and the internal resistance (r) in the circuit. i.e
R = + r
= R - r [Substitute the values of R and r]
= 5kΩ - 1kΩ
= 4kΩ
Therefore the series resistance is 4kΩ
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Answer:
1.
2.
3.The results from part 1 and 2 agree when r = R.
Explanation:
The volume charge density is given as
We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.
1. Since the cylinder is very long, Gauss’ Law can be applied.
The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is
where ‘h’ is the length of the imaginary Gaussian surface.
2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,
3. At the boundary where r = R:
As can be seen from above, two E-field values are equal as predicted.