In the given circuit, the current in $8 \; \Omega$ resistance is $1.5\text{ A}$. The total current ($I$) flowing in the circuit is
Show Hint
In parallel circuits, current distributes inversely to resistance: $\frac{I_2}{I_1} = \frac{R_1}{R_2}$. Here, the resistance ratio is $\frac{8}{3}$, so $I_2 = 1.5 \times \frac{8}{3} = 4\text{ A}$. Adding them up ($1.5 + 4$) instantly gives $5.5\text{ A}$.
Step 1: Understand the layout. An $8\ \Omega$ resistor and a $3\ \Omega$ resistor are connected in parallel. The $8\ \Omega$ branch carries $1.5\ \text{A}$. We want the total line current $I$ feeding the pair. Step 2: Key idea for parallel branches. Resistors in parallel share the same voltage across them. So if we find the voltage from one branch, it applies to the other too. Step 3: Find the common voltage. Using Ohm's law on the $8\ \Omega$ branch: $V = I_1 R_1 = 1.5 \times 8 = 12\ \text{V}$. Step 4: Find the current in the second branch. The $3\ \Omega$ branch also has $12\ \text{V}$ across it, so $I_2 = \dfrac{V}{R_2} = \dfrac{12}{3} = 4\ \text{A}$. Step 5: Apply Kirchhoff's current law. The total current entering the junction equals the sum of branch currents: $I = I_1 + I_2 = 1.5 + 4$. Step 6: Add them up. $I = 5.5\ \text{A}$. \[ \boxed{I = 5.5\ \text{A (option 4)}} \]
