The resistance of an ammeter of range 5A is `1.8Omega` A shunt of `0.2Omega` is connected in parallel to it. When its indicator shows a current of 2A then what will be effective current?

Options

A. The linear mass density continuously decreases from left to right.
B. The linear mass density continuously increases from left to right.
C. The linear mass density decreases from left to fight upto centre and then increases.
D. The linear mass density increases from left to right upto cente and then decreases.

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Identify the given values
- Resistance of the ammeter (R) = 1.8 Ω
- Resistance of the shunt (R_shunt) = 0.2 Ω
- Current shown by the ammeter (I) = 2 A

### Step 2: Calculate the voltage across the ammeter
Using Ohm's Law, we can calculate the voltage across the ammeter when it shows a current of 2 A:

\[ V = I \times R \]
\[ V = 2 \, \text{A} \times 1.8 \, \Omega = 3.6 \, \text{V} \]

### Step 3: Determine the current through the shunt
The voltage across the shunt is the same as the voltage across the ammeter, which is 3.6 V. Now, we can calculate the current through the shunt (I_shunt) using Ohm's Law:

\[ I_{shunt} = \frac{V}{R_{shunt}} \]
\[ I_{shunt} = \frac{3.6 \, \text{V}}{0.2 \, \Omega} = 18 \, \text{A} \]

### Step 4: Calculate the effective current
The effective current (I_E) in the circuit is the sum of the current through the ammeter and the current through the shunt:

\[ I_E = I + I_{shunt} \]
\[ I_E = 2 \, \text{A} + 18 \, \text{A} = 20 \, \text{A} \]

### Final Answer
The effective current is **20 A**.

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