Physics - Current Electricity Question with Solution | TestHub

The given circuit is a Wheatstone bridge.

Let the nodes be A (connected to the positive terminal), B (top junction), D (bottom junction), and C (connected to the negative terminal).

The resistances of the arms are $R_{AB}=6\Omega$, $R_{AD}=6\Omega$, $R_{BC}=4\Omega$, and $R_{DC}=4\Omega$. The resistance of the central arm is $R_{BD}=10\Omega$.

The ratio of resistances in the adjacent arms is:
$\frac{R_{AB}}{R_{AD}}=\frac{6}{6}=1$
$\frac{R_{BC}}{R_{DC}}=\frac{4}{4}=1$

Since $\frac{R_{AB}}{R_{AD}}=\frac{R_{BC}}{R_{DC}}$, the Wheatstone bridge is balanced.

Therefore, no current flows through the $10\Omega$ resistor, and it can be removed from the circuit.

The circuit simplifies to two parallel branches:
Resistance of the upper branch $=6\Omega +4\Omega =10\Omega$
Resistance of the lower branch $=6\Omega +4\Omega =10\Omega$

The equivalent resistance of the circuit is:
$R_{eq}=\frac{10\times 10}{10+10}=5\Omega$

The current drawn from the battery is:
$I=\frac{V}{R_{eq}}=\frac{12}{5}=2.4\text{ A}$

Answer: 2.4 A