Grade 12Electric Current

I modified the circuit as much as I can. Can I modify it any further?

Abhishek Das

9 Years agoGrade 12

9 Years ago

That modification is incorrect...

Starting from the right

$2\Omega + 1\Omega=3\Omega$ (Series Resistance)

${(\frac{1}{6}\Omega+\frac{1}{3}\Omega)}^{-1}=2\Omega$ (Parallel Resistance)

$2\Omega+2\Omega=4\Omega$ (Series Resistance)

${(\frac{1}{4}\Omega+\frac{1}{3}\Omega)}^{-1}=\frac{12}{7}\Omega$ (Parallel Resistance)

$\frac{12}{7}\Omega+4\Omega+4\Omega=\frac{68}{7}\Omega=9\frac{5}{7}\Omega \approx 9.7\Omega$ (Series Resistance)

Finaly the Circuit will be left with only one resistor of $9.7\Omega$ and a cell of $120V$

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