A shunt of resistance 1 $$\Omega$$ is connected across a galvanometer of 120 $$\Omega$$ resistance. A current of 5.5 ampere gives full scale deflection in the galvanometer. The current that will give full scale deflection in the absence of the shunt is nearly :

We are given a galvanometer with a resistance of 120 Ω and a shunt resistance of 1 Ω connected across it. The total current that produces full-scale deflection when the shunt is present is 5.5 amperes. We need to find the current that would cause full-scale deflection without the shunt. This current is the maximum current the galvanometer can handle on its own, denoted as $$ I_g $$.

When the shunt is connected, the total current $$ I_{\text{total}} = 5.5 $$ A splits into two parts: $$ I_g $$ through the galvanometer and $$ I_s $$ through the shunt. Since the shunt and galvanometer are in parallel, the voltage across them is the same. Therefore, we can write:

Voltage across galvanometer = Voltage across shunt

$$ I_g \times G = I_s \times S $$

where $$ G = 120 \Omega $$ and $$ S = 1 \Omega $$. Substituting the values:

$$ I_g \times 120 = I_s \times 1 $$

which simplifies to:

$$ I_s = 120 I_g \quad \text{(Equation 1)} $$

The total current is the sum of the currents through both paths:

$$ I_g + I_s = 5.5 \text{A} \quad \text{(Equation 2)} $$

Substitute $$ I_s $$ from Equation 1 into Equation 2:

$$ I_g + 120 I_g = 5.5 $$

Combine like terms:

$$ 121 I_g = 5.5 $$

Solve for $$ I_g $$:

$$ I_g = \frac{5.5}{121} $$

Convert 5.5 to a fraction: $$ 5.5 = \frac{11}{2} $$. So:

$$ I_g = \frac{\frac{11}{2}}{121} = \frac{11}{2 \times 121} = \frac{11}{242} $$

Simplify the fraction by dividing numerator and denominator by 11:

$$ \frac{11 \div 11}{242 \div 11} = \frac{1}{22} $$

Now, convert $$ \frac{1}{22} $$ to a decimal:

$$ \frac{1}{22} \approx 0.0454545\ldots $$

Rounding to three decimal places, $$ I_g \approx 0.045 $$ A.

This $$ I_g $$ is the current that causes full-scale deflection in the galvanometer when no shunt is present. Comparing with the options:

A. 5.5 ampere

B. 0.5 ampere

C. 0.004 ampere

D. 0.045 ampere

The value 0.045 A matches option D.

Hence, the correct answer is Option D.