1.

Find the current flowing through the 12 ohm resistor

Answer»

Explanation:

{eq}V = 21 \space V {/eq}

{eq}R_{1} = R_{3} = 3.8 \space \Omega {/eq}

{eq}R_{2} = R_{4} = 8.8 \space \Omega {/eq}

The CIRCUIT diagram can be simplified by taking the effective CAPACITANCE of {eq}R_{1}, R_{2}, {/eq} and {eq}R_{4} {/eq}. This is SIMPLY:

{eq}R_{1,2,4} = (\frac{1}{8.8} +\frac{1}{8.8})^{-1} + 3.8 {/eq}

{eq}R_{1,2,4} = 8.2 \space \Omega {/eq}

The other HALF of the resitors containing the {eq}12 \space \Omega {/eq} resistor:

{eq}R_{half} = (\frac{1}{4} +\frac{1}{12})^{-1} + 2.0 {/eq}

{eq}R_{half} = 5 \space \Omega {/eq}

The effective resistance of the whole circuit is therefore:

{eq}R_{eff} = (\frac{1}{5} +\frac{1}{8.2})^{-1} + 3.8 {/eq


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