Question

In the given figure, an ammeter $A$ consists of a $240 \, \Omega$ coil connected in parallel to a $10 \, \Omega$ shunt. The reading of the ammeter is ____ $\text{mA}$.

Answer 1

Correct Answer: 160

The problem asks for the reading of an ammeter in a simple series circuit. The ammeter is not ideal; it has an internal structure consisting of a coil and a shunt resistor connected in parallel. To find the ammeter's reading, which is the total current flowing through it, we must first calculate the equivalent resistance of the ammeter and then the total resistance of the entire circuit.

Concept Used:

The solution involves the application of Ohm's Law and the formulas for calculating equivalent resistance in series and parallel circuits.

1. Equivalent Resistance of Parallel Resistors: For two resistors, \(R_1\) and \(R_2\), connected in parallel, their equivalent resistance \(R_p\) is given by: \[ R_p = \frac{R_1 R_2}{R_1 + R_2} \]
2. Equivalent Resistance of Series Resistors: For resistors connected in series, the total resistance \(R_s\) is the sum of the individual resistances: \[ R_s = R_1 + R_2 + \ldots \]
3. Ohm's Law: The total current \(I\) flowing in a circuit is equal to the total voltage \(V\) divided by the total equivalent resistance \(R_{\text{total}}\): \[ I = \frac{V}{R_{\text{total}}} \]

The reading of an ammeter is the total current that enters the instrument.

Step-by-Step Solution:

Step 1: Calculate the equivalent resistance of the ammeter.

The ammeter consists of a coil with resistance \(R_{\text{coil}} = 240 \, \Omega\) connected in parallel with a shunt resistor \(R_{\text{shunt}} = 10 \, \Omega\). The equivalent resistance of the ammeter, \(R_A\), is:

\[ R_A = \frac{R_{\text{coil}} \times R_{\text{shunt}}}{R_{\text{coil}} + R_{\text{shunt}}} \] \[ R_A = \frac{240 \times 10}{240 + 10} = \frac{2400}{250} = \frac{240}{25} = 9.6 \, \Omega \]

Step 2: Calculate the total resistance of the circuit.

The ammeter is connected in series with an external resistor \(R_{\text{ext}} = 140.4 \, \Omega\). The total resistance of the circuit, \(R_{\text{total}}\), is the sum of the external resistance and the ammeter's resistance:

\[ R_{\text{total}} = R_{\text{ext}} + R_A \] \[ R_{\text{total}} = 140.4 \, \Omega + 9.6 \, \Omega = 150 \, \Omega \]

Step 3: Use Ohm's Law to find the total current in the circuit.

The voltage of the source is \(V = 24 \, \text{V}\). The total current \(I_{\text{total}}\) flowing through the circuit is what the ammeter measures.

\[ I_{\text{total}} = \frac{V}{R_{\text{total}}} \] \[ I_{\text{total}} = \frac{24 \, \text{V}}{150 \, \Omega} = 0.16 \, \text{A} \]

Step 4: Convert the current to milliamperes (mA).

The problem asks for the reading in mA. Since \(1 \, \text{A} = 1000 \, \text{mA}\), we convert the result:

\[ 0.16 \, \text{A} = 0.16 \times 1000 \, \text{mA} = 160 \, \text{mA} \]

The reading of the ammeter is 160 mA.

Answer 2

The circuit has a total resistance composed of a \( 140.4 \, \Omega \) resistor in series with the parallel combination of \( 240 \, \Omega \) and \( 10 \, \Omega \) resistors.
The equivalent resistance \( R_{\text{eq}} \) of the parallel combination is:
\[R_{\text{eq}} = \frac{240 \times 10}{240 + 10} = \frac{2400}{250} = 9.6 \, \Omega\]
Thus, the total resistance in the circuit becomes:
\[R_{\text{total}} = 140.4 + 9.6 = 150 \, \Omega\]
Now, the current \( I \) in the circuit is:
\[I = \frac{V}{R_{\text{total}}} = \frac{24}{150} = 0.16 \, \text{A} = 160 \, \text{mA}\]
Therefore, the current in the ammeter is \( 160 \, \text{mA} \).