A current through 1 ohm resistance in the following circuit is
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When checking if a bridge with decimals is balanced, multiply the numbers out to clear the decimals mentally: $\frac{12.5}{2.5}$ is exactly the same as $\frac{125}{25} = 5$. Spotting a balanced bridge instantly eliminates the need for complex Kirchhoff's Loop Law matrix systems!
Step 1: Understand the question. We must find the current through the $1\ \Omega$ resistor in the given network using Kirchhoff's circuit laws. The idea is to assign loop currents and write voltage equations. Step 2: State the two laws. Kirchhoff's current law says the total current entering a junction equals the total leaving. Kirchhoff's voltage law says the sum of voltage changes around any closed loop is zero. Step 3: Assign currents. Label the branch currents with symbols, for example $I_1$ and $I_2$ in the two loops, and let the current through the $1\ \Omega$ resistor be the combination of these as set by the junction rule. Step 4: Write the loop equations. Going around each closed loop, add up the emf sources and subtract the drops $IR$ across each resistor, and set the total to zero. This gives two equations in $I_1$ and $I_2$. Step 5: Solve the equations. Solving the pair of simultaneous equations gives the branch currents. Substituting back, the current passing through the $1\ \Omega$ resistor comes out to $1.8$ A. Step 6: State the answer. The current through the $1\ \Omega$ resistance is $1.8$ A. \[ \boxed{I_{1\Omega} = 1.8\ \text{A}} \]
