A galvanometer has a resistance of 100?. A resistance of 1? is connected across its terminal. What part of total current flows through the galvanometer?

To determine what part of the total current flows through the galvanometer when a resistance is connected across its terminals, we can use the concept of parallel circuits. In this case, we have the galvanometer and the additional resistance connected in parallel, which affects how the current is divided between them.

Understanding the Circuit Configuration

In a parallel circuit, the voltage across each component is the same. The total current flowing through the circuit is the sum of the currents flowing through each branch. Here, we have:

• Resistance of the galvanometer (Rg) = 100 Ω
• Resistance connected across the galvanometer (R) = 1 Ω

Calculating Total Resistance

The total resistance (Rt) of two resistors in parallel can be calculated using the formula:

1/Rt = 1/Rg + 1/R

Substituting the values:

1/Rt = 1/100 + 1/1

This simplifies to:

1/Rt = 0.01 + 1 = 1.01

Now, taking the reciprocal gives us:

Rt = 1 / 1.01 ≈ 0.9901 Ω

Finding Current Distribution

Next, we can find the total current (I) flowing through the circuit using Ohm's Law, which states:

I = V / R

Assuming we have a voltage (V) applied across the parallel combination, the total current can be expressed as:

I = V / Rt

Current Through Each Resistor

Now, we can find the current through each resistor. The current through the galvanometer (Ig) can be calculated using:

Ig = V / Rg

And the current through the 1 Ω resistor (I1) is:

I1 = V / R

Current Ratio

The ratio of the currents can be expressed as:

Ig / I1 = (V / Rg) / (V / R) = R / Rg

Substituting the values:

Ig / I1 = 1 / 100

This means that for every 1 part of current flowing through the galvanometer, 100 parts flow through the 1 Ω resistor.

Final Calculation of Current Through the Galvanometer

To find the fraction of the total current that flows through the galvanometer, we can express it as:

Ig / Itotal = 1 / (1 + 100) = 1 / 101

This indicates that approximately 0.9901% of the total current flows through the galvanometer, while the vast majority flows through the 1 Ω resistor.

In summary, when a 1 Ω resistor is connected across a galvanometer with a resistance of 100 Ω, only a small fraction of the total current flows through the galvanometer, illustrating the principles of parallel circuits and current division.