To find the total current drawn from the battery, when two resistors are connected in parallel, we first need to calculate the equivalent resistance of the parallel combination of resistors.
For resistors in parallel, the equivalent resistance \( R_{\text{eq}} \) is given by the formula:
\( \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} \)
Substituting the given resistor values, \( R_1 = 4 \Omega \) and \( R_2 = 6 \Omega \):
\( \frac{1}{R_{\text{eq}}} = \frac{1}{4} + \frac{1}{6} \)
To add these fractions, we find a common denominator, which is 12:
\( \frac{1}{4} = \frac{3}{12}, \quad \frac{1}{6} = \frac{2}{12} \)
\( \frac{1}{R_{\text{eq}}} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12} \)
Thus:
\( R_{\text{eq}} = \frac{12}{5} \Omega = 2.4 \Omega \)
Now, using Ohm's law, the total current \( I \) drawn from the battery is given by:
\( I = \frac{V}{R_{\text{eq}}} \)
where \( V = 12 \text{ V} \):
\( I = \frac{12 \text{ V}}{2.4 \Omega} = 5 \text{ A} \)
Therefore, the total current drawn from the battery is 5 A.

The equivalent resistance between the points \(A\) and \(B\) in the given circuit is \[ \frac{x}{5}\,\Omega. \] Find the value of \(x\). 
A Wheatstone bridge is initially at room temperature and all arms of the bridge have same value of resistances \[ (R_1=R_2=R_3=R_4). \] When \(R_3\) resistance is heated, its resistance value increases by \(10%\). The potential difference \((V_a-V_b)\) after \(R_3\) is heated is _______ V. 
The heat generated in 1 minute between points A and B in the given circuit, when a battery of 9 V with internal resistance of 1 \(\Omega\) is connected across these points is ______ J. 