A galvanometer having a resistance of 8 ohm is shunted by a wire of resistance 2 ohm . If the total current is 1 amp , the part of it passing through the shunt will be

Options

A. 0.5 A
B. 0.2 A
C. 0.8 A
D. 0.25 A

Answer

To solve the problem step by step, we need to analyze the circuit involving the galvanometer and the shunt resistor. Here’s how we can approach it:

### Step 1: Understand the Circuit
We have a galvanometer with a resistance \( R_g = 8 \, \Omega \) and a shunt resistor with a resistance \( R_s = 2 \, \Omega \). The total current flowing into the circuit is \( I = 1 \, \text{A} \).

### Step 2: Set Up the Current Division
Let \( I_s \) be the current flowing through the shunt resistor. The current flowing through the galvanometer will then be \( I_g = I - I_s \).

### Step 3: Apply the Concept of Parallel Resistors
Since the galvanometer and the shunt resistor are in parallel, the voltage across both resistors is the same. Therefore, we can write the following equation based on Ohm's Law:

\[ I_s \cdot R_s = (I - I_s) \cdot R_g \]

### Step 4: Substitute the Known Values
Substituting the known values into the equation:

\[ I_s \cdot 2 = (1 - I_s) \cdot 8 \]

### Step 5: Expand and Rearrange the Equation
Expanding the equation gives:

\[ 2 I_s = 8 - 8 I_s \]

Now, rearranging the equation to isolate \( I_s \):

\[ 2 I_s + 8 I_s = 8 \]

\[ 10 I_s = 8 \]

### Step 6: Solve for \( I_s \)
Now, divide both sides by 10:

\[ I_s = \frac{8}{10} = 0.8 \, \text{A} \]

### Conclusion
The part of the total current passing through the shunt resistor is \( 0.8 \, \text{A} \).

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