Identify X and y in the circuit diagram. Calculate the total resistance, the reading of X, and explain why X is connected there. Also, explain the effect of removing the 2 ohm resistor on the current.

Step 1: Identify X and Y
X is an ammeter.
Y is a voltmeter.

Step 2: Calculate the total resistance of the circuit
Each branch has resistors in series:
Upper branch: $4\,\Omega + 1\,\Omega = 5\,\Omega$
Lower branch: $4\,\Omega + 1\,\Omega = 5\,\Omega$
The voltmeter Y has infinite resistance and since $\frac{4}{4} = \frac{1}{1}$, the bridge is balanced, so no current flows through Y.
The two branches are in parallel:
$\frac{1}{R_total} = \frac{1}{5\,\Omega} + \frac{1}{5\,\Omega} = \frac{2}{5}\,\Omega^{-1}$
$R_total = \frac{5}{2} = 2.5\,\Omega$
2.5 \Omega

Step 3: Calculate the reading of I
I is the total current through the ammeter X.
$I = \frac{V}{R_total} = \frac{24\,V}{2.5\,\Omega}$
$24 \div 2.5 = 9.6\,\mathrm{A}$
9.6 A

Step 4: Justify why X is connected in the position in the circuit
X (ammeter) measures the total current supplied by the battery. It must be connected in series between the battery and the parallel branches so that the entire current passes through it. An ammeter has very low resistance, so it does not significantly change the current in the circuit.

Step 5: State and explain what will happen to the value of the current in the circuit if the 2nd resistor is removed
The 2nd resistor refers to one of the $1\,\Omega$ resistors (e.g., lower $1\,\Omega$).
State: The current will decrease.
Explain: Removing the $1\,\Omega$ resistor opens that parallel branch (equivalent resistance becomes infinite).
Only one branch remains: $R'_total = 5\,\Omega$.
$I' = \frac{24\,V}{5\,\Omega} = 4.8\,\mathrm{A}$
Since total resistance increases from $2.5\,\Omega$ to $5\,\Omega$, by Ohm's law ($I = \frac{V}{R}$), the current decreases for constant voltage.