Physics - Current Electricity Question with Solution | TestHub

Let the nodes of the circuit be labeled as A, B, C, and D, where A is connected to the positive terminal and D to the negative terminal of the battery.

From the circuit diagram, the resistances between the nodes are:
$R_{AB}=5\Omega$
$R_{AC}=10\Omega$
$R_{BD}=10\Omega$
$R_{CD}=20\Omega$
$R_{BC}=10\Omega$

This forms a Wheatstone bridge. Checking the balance condition:
$\frac{R_{AB}}{R_{BD}}=\frac{5}{10}=\frac{1}{2}$
$\frac{R_{AC}}{R_{CD}}=\frac{10}{20}=\frac{1}{2}$

Since $\frac{R_{AB}}{R_{BD}}=\frac{R_{AC}}{R_{CD}}$, the bridge is balanced. Thus, no current flows through the central resistor $R_{BC}=10\Omega$, and it can be removed from the circuit.

The equivalent resistance $R_{eq}$ is the parallel combination of the series branches A-B-D and A-C-D:
$R_{ABD}=5+10=15\Omega$
$R_{ACD}=10+20=30\Omega$

$\frac{1}{R_{eq}}=\frac{1}{15}+\frac{1}{30}=\frac{3}{30}=\frac{1}{10}{\Omega }^{-1}$
$\Rightarrow R_{eq}=10\Omega$

The current $I$ drawn from the battery is:
$I=\frac{V}{R_{eq}}=\frac{5}{10}=0.5\text{ A}$

Answer: $0.5\mathrm{A}$