The effect of closing the key K in the circuit shown above would be to
A.
increase the current by 0.6A
B.
reduce the current by 0.6A
C.
increase the current by 0.4A
D.
keep the current unchanged
Correct answer is C
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