Question

In the figure shown below, a resistance of 150.4 $ \Omega $ is connected in series to an ammeter A of resistance 240 $ \Omega $. A shunt resistance of 10 $ \Omega $ is connected in parallel with the ammeter. The reading of the ammeter is ______ mA.

Hint:

Calculate the equivalent resistance of the circuit. Then use Ohm's law and current division to find the current through the ammeter.

Answer 1

Correct Answer: 5

This problem involves a circuit with a voltage source, a series resistor, and an ammeter shunted by another resistor. We need to find the reading of the ammeter, which is the current flowing through it.

Concept Used:

The solution requires the application of several fundamental circuit laws:

1. Equivalent Resistance in Parallel: For two resistors \( R_1 \) and \( R_2 \) in parallel, the equivalent resistance \( R_p \) is given by: \[ R_p = \frac{R_1 R_2}{R_1 + R_2} \]
2. Equivalent Resistance in Series: For resistors in series, the total resistance is the sum of individual resistances.
3. Ohm's Law: The total current \( I \) flowing in a circuit is given by \( I = \frac{V}{R_{eq}} \), where \( V \) is the total voltage and \( R_{eq} \) is the total equivalent resistance of the circuit.
4. Current Divider Rule: When a total current \( I_{total} \) enters a parallel combination of two resistors \( R_1 \) and \( R_2 \), the current \( I_1 \) through resistor \( R_1 \) is: \[ I_1 = I_{total} \times \frac{R_2}{R_1 + R_2} \]

 

Step-by-Step Solution:

Step 1: Calculate the equivalent resistance of the parallel combination of the ammeter and the shunt resistor.

The ammeter has a resistance \( R_A = 240 \, \Omega \), and the shunt resistor has a resistance \( R_{sh} = 10 \, \Omega \). These two are connected in parallel. \[ R_p = \frac{R_A \times R_{sh}}{R_A + R_{sh}} \] \[ R_p = \frac{240 \times 10}{240 + 10} = \frac{2400}{250} = 9.6 \, \Omega \]

Step 2: Calculate the total equivalent resistance of the entire circuit.

The series resistor \( R_s = 150.4 \, \Omega \) is connected in series with the parallel combination \( R_p \). \[ R_{eq} = R_s + R_p \] \[ R_{eq} = 150.4 + 9.6 = 160 \, \Omega \]

Step 3: Calculate the total current flowing from the voltage source.

The voltage of the source is \( V = 20 \, \text{V} \). Using Ohm's Law: \[ I_{total} = \frac{V}{R_{eq}} = \frac{20}{160} = \frac{1}{8} \, \text{A} \] \[ I_{total} = 0.125 \, \text{A} \]

Step 4: Calculate the current flowing through the ammeter using the current divider rule.

The total current \( I_{total} \) splits between the ammeter (\( R_A \)) and the shunt resistor (\( R_{sh} \)). The reading of the ammeter is the current flowing through it, \( I_A \). \[ I_A = I_{total} \times \frac{R_{sh}}{R_A + R_{sh}} \] \[ I_A = 0.125 \times \frac{10}{240 + 10} \] \[ I_A = 0.125 \times \frac{10}{250} = 0.125 \times \frac{1}{25} \] \[ I_A = \frac{0.125}{25} = 0.005 \, \text{A} \]

Step 5: Convert the ammeter reading to milliamperes (mA).

To convert from amperes to milliamperes, we multiply by 1000. \[ I_A = 0.005 \, \text{A} \times 1000 \, \frac{\text{mA}}{\text{A}} = 5 \, \text{mA} \]

The reading of the ammeter is 5 mA.

Answer 2

$R_{eq} = R_1 + R_2$

$R_{eq} = 150.4 + \frac{240 \times 10}{250}$

$R_{eq} = 150.4 + 9.6 = 160 \Omega$

$I_1 = \frac{IR_2}{240}$

$I_1 = \frac{I \times 9.6}{240}$

$I = \frac{20}{160}$

$I_1 = \frac{20}{160} \times \frac{9.6}{240} = \frac{1}{200} = 5 \times 10^{-3} A = 5 mA$