In the given circuit, the current in $8 \; \Omega$ resistance is $1.5\text{ A}$. The total current ($I$) flowing in the circuit is
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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: Understanding the Question:
We are given a parallel resistor network. The branch containing an $8 \; \Omega$ resistor carries a known current of $1.5\text{ A}$. Another branch connected in parallel with it contains a $3 \; \Omega$ resistor. We need to find the total combined line current ($I$) feeding this parallel assembly.
Step 2: Key Formula or Approach:
For components connected in parallel, the potential difference ($V$) across each parallel branch is exactly identical:
$$V = I_1 R_1 = I_2 R_2$$
According to Kirchhoff's Current Law (KCL), the total current entering a junction must equal the sum of the currents branches splitting from that junction:
$$I = I_1 + I_2$$
Step 3: Detailed Explanation:
Let the first branch have resistance $R_1 = 8 \; \Omega$ and current $I_1 = 1.5\text{ A}$.
Calculate the voltage drop across this branch using Ohm's Law:
$$V = I_1 \cdot R_1 = 1.5\text{ A} \times 8 \; \Omega = 12\text{ V}$$
Since the second branch ($R_2 = 3 \; \Omega$) is in parallel, it experiences the exact same voltage drop of $12\text{ V}$. Let's calculate its branch current $I_2$:
$$I_2 = \frac{V}{R_2} = \frac{12\text{ V}}{3 \; \Omega} = 4\text{ A}$$
Now, sum the individual branch currents together to determine the total input line current $I$:
$$I = I_1 + I_2 = 1.5\text{ A} + 4\text{ A} = 5.5\text{ A}$$
Step 4: Final Answer:
The total current flowing in the circuit is $5.5\text{ A}$, which matches option (D).
