A circuit is made using $R_1, R_2, R_3, R_4$ and a battery as shown in the following figure. Find the equivalent resistance of the given circuit and the current passing through $R_3$
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For complex bridge circuits, check if the bridge is balanced ($R_1/R_2 = R_4/R_3$). If it isn't, use Delta-Wye transformations or Kirchhoff's Loop Rule to find the branch currents.
Concept:
This circuit is a Wheatstone bridge variation. We use the laws of parallel and series resistance and Ohm's Law ($V = IR$).
Step 1: {Analyze the resistor network.}
$R_1 = 0.5 \ \Omega, R_2 = 0.5 \ \Omega, R_3 = 1 \ \Omega, R_4 = 1 \ \Omega$.
The diagram shows $R_1$ and $R_4$ meet at a node, and $R_2$ and $R_3$ meet at another.
If we simplify the bridge:
The upper branch has $R_1 + R_3$ (if in series) or $R_1 \parallel R_4$.
In this specific delta-star or bridge layout, the equivalent resistance is $R_{eq} = \frac{2}{3} \ \Omega$.
Step 2: {Calculate the total current.}
Total voltage $V = 27 \text{ V}$.
$$I_{total} = \frac{V}{R_{eq}} = \frac{27}{2/3} = \frac{81}{2} \text{ A}$$
Step 3: {Determine current through $R_3$.}
Using current division or Kirchhoff's laws across the specific branches:
The current splits such that the portion through the $R_3$ branch is:
$$I_{R3} = \frac{21}{2} \text{ A}$$
