Question 80: Current Electricity

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 .

Accepted Answer

Step 1: Find the Total Resistance
First, add up the resistor in series ($R_1 = 150.4~\Omega$) with the combined resistance of the ammeter and its shunt ($R_2$).
The ammeter ($240~\Omega$) is in parallel with a shunt of ($10~\Omega$). The combined resistance of resistors in parallel is found by:
$$ R_2 = \frac{240 	imes 10}{240 + 10} = \frac{2400}{250} = 9.6~\Omega $$
The total resistance ($R_{eq}$) is:
$$ R_{eq} = 150.4 + 9.6 = 160~\Omega $$
Step 2: Find the Current in the Circuit
Suppose the total current in the circuit is $I$. But to find the ammeter reading, we care about the current through the ammeter coil itself.
The current through the ammeter, $I_1$, is given by how the current splits between the ammeter and the shunt. The fraction going through the ammeter is:
$$ I_1 = \frac{R_2}{240} 	imes I $$
But $R_2 = 9.6~\Omega$ (the parallel combination), and $240~\Omega$ is the ammeter's resistance:
$$ I_1 = \frac{9.6}{240} 	imes I $$
Step 3: Substitute the Circuit Values
The supply voltage is $20~V$. The total resistance is $160~\Omega$. So the total current is:
$$ I = \frac{20}{160} = 0.125~A $$
So:

$$ I_1 = 0.125 	imes \frac{9.6}{240} $$
Calculate this:

$$ I_1 = 0.125 	imes 0.04 = 0.005~A = 5~	ext{mA} $$
Final Answer:
The ammeter reads $5~	ext{mA}$.

Current Electricity

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 .

Integer / Numeric Answer Type