The inductance of a moving iron ammeter with a full-scale deflection of 900 at 1.5A, is given by the expression 𝐿 = 200 + 40θ − 4θ^2 - θ^3 µ𝐻, where θ is deflection in radian from the zero position. Estimate the angular deflection of the pointer for a current of 1.0A.

To estimate the angular deflection \(\theta\) of the pointer for a current of 1.0 A, we first need to understand the relationship between the current and the deflection of the ammeter. The energy stored in the inductance of the ammeter is responsible for the deflection of the pointer. For a moving iron instrument, the torque is proportional to the rate of change of inductance with respect to the deflection.

### Step 1: Expression for the inductance

The inductance of the ammeter is given by:

\[
L(\theta) = 200 + 40\theta - 4\theta^2 - \theta^3 \ \mu H
\]

### Step 2: Energy stored in the inductance

The energy stored in an inductor \( L \) with current \( I \) is given by:

\[
W = \frac{1}{2} I^2 L
\]

The deflection torque (\( T \)) is proportional to the rate of change of energy with respect to deflection:

\[
T = \frac{dW}{d\theta} = \frac{1}{2} I^2 \frac{dL}{d\theta}
\]

To find \(\frac{dL}{d\theta}\):

\[
\frac{dL}{d\theta} = \frac{d}{d\theta} (200 + 40\theta - 4\theta^2 - \theta^3) = 40 - 8\theta - 3\theta^2
\]

So, the torque \( T \) can be expressed as:

\[
T = \frac{1}{2} I^2 (40 - 8\theta - 3\theta^2)
\]

### Step 3: Full-scale deflection condition

Given that full-scale deflection of 900 (which is \(\frac{\pi}{2}\) radians) occurs at 1.5 A, we can set up an equation for the torque at full scale:

\[
T_{fs} = \frac{1}{2} (1.5)^2 \left( 40 - 8 \cdot \frac{\pi}{2} - 3 \left(\frac{\pi}{2}\right)^2 \right)
\]

This will give us a reference torque value \( T_{fs} \).

### Step 4: Find torque at 1.0 A and equate to the reference torque

We set up the equation for the torque at 1.0 A:

\[
T_{1.0} = \frac{1}{2} (1.0)^2 (40 - 8\theta - 3\theta^2)
\]

Since \( T_{1.0} \) is proportional to \( T_{fs} \), we can solve for \(\theta\). Equate the ratio of torques to the ratio of currents squared:

\[
\frac{T_{1.0}}{T_{fs}} = \left(\frac{1.0}{1.5}\right)^2
\]

Substituting \( T_{1.0} \) and \( T_{fs} \), and then solving for \(\theta\), let's proceed with the calculations.The possible solutions for the angular deflection \(\theta\) are:

\[
\theta \approx -4.24 \text{ radians (not feasible since it's negative)}
\]
\[
\theta \approx 1.57 \text{ radians}
\]

The feasible solution is \(\theta \approx 1.57\) radians, which is equivalent to approximately 90 degrees. This shows that the ammeter's deflection is quite significant at 1.0 A, nearly reaching the full-scale deflection point.

Would you like further clarification, or do you have any additional questions?

### Related Questions
1. How does the inductance affect the deflection in a moving iron instrument?
2. What is the significance of the negative solution in this context?
3. How would the deflection change if the current was halved to 0.5 A?
4. Can the relationship between current and deflection be considered linear based on the given expression?
5. How would you determine the deflection if the inductance formula had a different form?

**Tip:** Always check the feasibility of solutions in real-world scenarios, especially when dealing with physical quantities like angles or distances.

This problem explores the estimation of angular deflection for a moving iron ammeter given the inductance expression L(θ) = 200 + 40θ − 4θ^2 − θ^3 µH and a current of 1.0A. Learn how to apply energy and torque formulas to solve for deflection, using principles of differentiation and inductance. Suitable for advanced high school and college-level physics students.

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