Question

A moving coil instrument gives full scale deflection of 24 mA. When a P.D. across if is 108 mV. Find the value of – (i) Series resistance for full scale deflection of 400 V. (ii) Find the power consumption.

Answer 1

To solve this problem, we need to find the series resistance required for full-scale deflection at a higher voltage and the power consumption of the moving coil instrument.

### Given Data:
- Full scale deflection current, \( I_f = 24 \, \text{mA} = 0.024 \, \text{A} \)
- Voltage across the instrument at full scale deflection, \( V_f = 108 \, \text{mV} = 0.108 \, \text{V} \)
- Desired full scale deflection voltage, \( V_d = 400 \, \text{V} \)

### Step 1: Calculate the internal resistance of the instrument.

Using Ohm's law, we can find the internal resistance \( R_m \) of the moving coil instrument:

\[
R_m = \frac{V_f}{I_f} = \frac{0.108 \, \text{V}}{0.024 \, \text{A}} = 4.5 \, \Omega
\]

### Step 2: Calculate the series resistance for full scale deflection at 400 V.

To achieve full-scale deflection of 400 V, we need to find the total resistance \( R_{total} \) required:

\[
R_{total} = \frac{V_d}{I_f} = \frac{400 \, \text{V}}{0.024 \, \text{A}} = 16666.67 \, \Omega
\]

The series resistance \( R_s \) required can be found using the formula:

\[
R_s = R_{total} - R_m
\]

Substituting the values:

\[
R_s = 16666.67 \, \Omega - 4.5 \, \Omega \approx 16662.17 \, \Omega
\]

### Step 3: Calculate the power consumption.

The power consumption \( P \) can be calculated using the formula:

\[
P = V_f \times I_f
\]

Substituting the values:

\[
P = 0.108 \, \text{V} \times 0.024 \, \text{A} = 0.002592 \, \text{W} = 2.592 \, \text{mW}
\]

### Summary of Results:
(i) Series resistance for full-scale deflection of 400 V: **16662.17 Ω**  
(ii) Power consumption: **2.592 mW**

Answer 2

To solve this problem, we'll need to follow these steps:

1. **Calculate the series resistance required for full-scale deflection of 400 V.**

2. **Determine the power consumption of the moving coil instrument.**

Let's break it down step by step:

### Given Data
- **Full-scale deflection current (I_fsd):** 24 µA (24 × 10⁻⁶ A)
- **Full-scale deflection voltage (V_fsd):** 108 mV (108 × 10⁻³ V)
- **Required full-scale deflection voltage (V_req):** 400 V

### (i) Finding the Series Resistance

1. **Calculate the internal resistance of the instrument:**

   The internal resistance \( R_{int} \) can be calculated using Ohm's Law:
   \[
   R_{int} = \frac{V_{fsd}}{I_{fsd}}
   \]

   Substituting the given values:
   \[
   R_{int} = \frac{108 \times 10^{-3} \text{ V}}{24 \times 10^{-6} \text{ A}} = \frac{108 \text{ mV}}{24 \text{ µA}} = 4.5 \text{ kΩ}
   \]

2. **Determine the series resistance needed for full-scale deflection at 400 V:**

   For the instrument to give full-scale deflection at 400 V, a series resistance \( R_s \) is required. The total resistance in the circuit should be such that the voltage across the instrument is 108 mV when the voltage across the series resistance is the remaining part of 400 V.

   So, the voltage across the series resistance \( V_s \) is:
   \[
   V_s = V_{req} - V_{fsd} = 400 \text{ V} - 108 \text{ mV} = 399.892 \text{ V}
   \]

   The current through the series resistor is the same as the current through the meter (full-scale deflection current):
   \[
   I = I_{fsd} = 24 \text{ µA}
   \]

   Therefore, the series resistance \( R_s \) can be calculated using Ohm's Law:
   \[
   R_s = \frac{V_s}{I_{fsd}} = \frac{399.892 \text{ V}}{24 \times 10^{-6} \text{ A}} = 16,666,333 \text{ Ω} \approx 16.67 \text{ MΩ}
   \]

### (ii) Finding the Power Consumption

1. **Calculate the power consumed by the instrument:**

   Power consumption \( P_{inst} \) of the moving coil instrument when it is giving full-scale deflection is:
   \[
   P_{inst} = V_{fsd} \times I_{fsd}
   \]

   Substituting the values:
   \[
   P_{inst} = 108 \times 10^{-3} \text{ V} \times 24 \times 10^{-6} \text{ A} = 2.592 \times 10^{-9} \text{ W} = 2.592 \text{ nW}
   \]

2. **Calculate the power consumed by the series resistance:**

   The power \( P_s \) consumed by the series resistance is:
   \[
   P_s = I_{fsd}^2 \times R_s
   \]

   Substituting the values:
   \[
   P_s = (24 \times 10^{-6} \text{ A})^2 \times 16,666,333 \text{ Ω} = 2.592 \times 10^{-9} \text{ W} \approx 2.592 \text{ nW}
   \]

### Summary

1. **Series Resistance Required:**
   - \( R_s \approx 16.67 \text{ MΩ} \)

2. **Power Consumption:**
   - The power consumed by the instrument is approximately \( 2.592 \text{ nW} \).

   Note: The power consumed by the series resistance is effectively the same as the power consumed by the instrument, given that the series resistance is designed to drop the excess voltage while the instrument measures the desired voltage.