Correct Answer: Option C

Explanation

When K is open, the total resistance

R = 8\(\Omega\) + 2\(\Omega\) = 10\(\Omega\)

I = \(\frac{V}{R}\) = \(\frac{6}{10}\)

= 0.6A

When K is closed, we have

\(\frac{1}{R}\) = \(\frac{1}{8}\) + \(\frac{1}{8}\)

⇒ \(\frac{1}{R}\) = \(\frac{1}{4}\)

⇒ R = 4\(\Omega\)

The total resistance R = 4\(\Omega\) + 2\(\Omega\)

= 6\(\Omega\)

I = \(\frac{6}{6}\)

= 1A

Hence, I = 1A - 0.6A = 0.4A increase

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