P 3.6-3 Find i using appropriate circuit reductions and the current divider principle for the circuit of Figure P 3.6-3.

Figure P 3.6-3

Starting at the far end away from the power source, we have two 1 Ω resistors in series which are effectively a 2 Ω resistor which is then in parallel with another 2 Ω resistor. Two 2 Ω resistors in parallel are equivalently a 1 Ω resistance, so then we are in series with another 1 Ω resistor so the pattern continues until we have a 12 V supply in series with a single 2Ω resistor. So the total current is

Now as we hit each branch we have 2 Ω resistance in parallel with another 2Ω resistance so the current splits in half at each one. So we have three splits so the current is reduced by

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This page last updated on January 20, 2019