Physics - Current Electricity Question with Solution | TestHub

The given circuit is a Wheatstone bridge.

Checking the balance condition:
$\frac{R_{AB}}{R_{BC}}=\frac{15}{3}=5$
$\frac{R_{AD}}{R_{DC}}=\frac{20}{4}=5$

Since $\frac{R_{AB}}{R_{BC}}=\frac{R_{AD}}{R_{DC}}$, the bridge is balanced and no current flows through the $6\Omega$ resistor in branch BD.

The circuit simplifies to two parallel branches:
Resistance of the upper branch, $R_1=15+3=18\Omega$
Resistance of the lower branch, $R_2=20+4=24\Omega$

The total current is $I=2.1\text{A}$.

Using the current divider rule, the current through the $15\Omega$ resistance (upper branch) is:
$I_{15\Omega }=I\times \frac{R_2}{R_1+R_2}$

$I_{15\Omega }=2.1\times \frac{24}{18+24}$

$I_{15\Omega }=2.1\times \frac{24}{42}=1.2\text{A}$

Answer: $1.2$ A