Question

In the following circuit, the current through \(6 \Omega\) resistor is

• A. \(\frac{1}{5}\text{ }\mathrm{A}\)
• B. \(\frac{2}{5}\text{ }\mathrm{A}\)
• C. \(\frac{1}{4}\text{ }\mathrm{A}\)
• D. \(\frac{3}{4}\text{ }\mathrm{A}\)

Hint:

The two cells face the same way and are in parallel, so replace them with one equivalent cell and combine resistances.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Both 3 V cells have their positive plates on the left side, so they are connected in parallel with the same polarity.
Each cell has a 3 \(\Omega\) resistor in its branch.

Step 2: Equivalent Source
Two equal cells in parallel give an emf of 3 V.
The two 3 \(\Omega\) resistors in parallel give \(\dfrac{3\times 3}{3+3} = 1.5\,\Omega\).

Step 3: Total Circuit
The equivalent source drives current through \(1.5 + 6 = 7.5\,\Omega\).

Step 4: Calculation
\(I = \dfrac{3}{7.5} = \dfrac{2}{5}\ \text{A}\)
This total current flows through the 6 \(\Omega\) resistor.

Final Answer:
The current in the 6 ohm resistor is two fifths of an ampere. \[ \boxed{\text{(B) }\dfrac{2}{5}\text{ A}} \]