Shunt Resistance Calculation for Doubling Current Range

This question asks us to determine the value of a shunt resistor needed to increase the current measuring capability of a meter movement. We are given the characteristics of the meter movement and the desired outcome.

Understanding Meter Shunts

An ammeter is used to measure electric current. A basic ammeter consists of a meter movement (often a galvanometer) that deflects proportionally to the current passing through it. However, the meter movement itself can typically only handle a very small current (its full-scale deflection current, $I_m$).

To measure larger currents, a low-resistance resistor, called a shunt resistor ($R_{sh}$), is connected in parallel with the meter movement ($R_m$). This arrangement allows the majority of the current to bypass the sensitive meter movement, protecting it and extending the instrument's range.

The key principle is that the voltage across the parallel components (meter movement and shunt) is the same. The total current ($I_{total}$) entering the parallel combination divides between the meter movement ($I_m$) and the shunt ($I_{sh}$), such that:

Since the voltage across both is equal:

From this, we can derive the formula for the shunt resistance:

Alternatively, substituting $I_{sh} = I_{total} - I_m$:

Calculation Steps

Let's break down the calculation based on the provided information:

• Given Meter Movement Current ($I_m$): The meter movement rating is 100 micro Amperes ($\mu A$). This is the maximum current the meter itself can handle for full-scale deflection. $I_m = 100 \, \mu A = 100 \times 10^{-6} \, A$
• Given Meter Movement Resistance ($R_m$): The resistance of the meter movement is 3000 Ohms ($\Omega$). $R_m = 3000 \, \Omega$
• Desired Current Range: The goal is to double the current range. This means the new total current capacity ($I_{total}$) should be twice the original meter movement current ($I_m$). $I_{total} = 2 \times I_m = 2 \times 100 \, \mu A = 200 \, \mu A$
• Calculate Shunt Current ($I_{sh}$): The current that flows through the shunt resistor is the total current minus the current flowing through the meter movement. $I_{sh} = I_{total} - I_m$ $I_{sh} = 200 \, \mu A - 100 \, \mu A = 100 \, \mu A$ $I_{sh} = 100 \times 10^{-6} \, A$
• Calculate Shunt Resistance ($R_{sh}$): Now, we use the derived formula: $R_{sh} = \frac{I_m \times R_m}{I_{sh}}$ Substitute the values: $R_{sh} = \frac{(100 \times 10^{-6} \, A) \times 3000 \, \Omega}{100 \times 10^{-6} \, A}$ The term '$100 \times 10^{-6} \, A$' cancels out in the numerator and the denominator. $R_{sh} = 3000 \, \Omega$

Final Answer Explanation

The calculation shows that the required shunt resistance is 3,000 Ohms. This value ensures that when the total current is 200 $\mu A$, 100 $\mu A$ goes through the meter movement (causing full-scale deflection) and the remaining 100 $\mu A$ goes through the shunt resistor.

Therefore, a 3,000 ohm shunt resistor is needed to double the current range of the 100 micro Amp, 3000-ohm meter movement.