The effect of closing the key K in the circuit shown above would be to
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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
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