WAEC - Physics (2000 - No. 35)
In the diagram above, the current passing through the 6\(\Omega\) resistor is 1.5A. Calculate the current in the 3\(\Omega\) resistor
1.20A
0.90A
0.75A
0.60A
Explanation
\(\frac{1}{\text{R}}\) = \(\frac{1}{2}\) + \(\frac{1}{3}\) = \(\frac{5}{6}\)
R = \(\frac{6}{5}\) = 1.2\(\Omega\); So p.d through the 3\(\Omega\) and 2\(\Omega\) resistors in parallel = 1.5 x 1.2 =1.8V
So the current through the 3\(\Omega\) resistor(I) = \(\frac{\text{V}}{\text{R}}\) = \(\frac{1.8}{3}\) = 0.6A (recall V = IR)
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