Question

Please use the Reduce and Return method to solve and show steps. thank you

Calculate R1, I1, Vo and Vab in the following circuit. (2..801kΩ, 5.287mA, - 10.415V, 14.27V) b 2.7kΩ 1.8kΩ 8.2kΩ 18V 1.5kΩ 1.2kΩ R=5.6kΩ a w

Answer 1

Reducing the resistance first 1.8 K and 2.7 K are in series so the total resistance is 4.5 K ohms and 1.5 K and 1.2 K resistance are also in series to each other so the combination gives 2.7 K ohms

now the combination of (1.8K+2.7K) and 8.2 K ohms resistors are in parallel so the combination gives
4.5 X 8.2 4.5 + 8.2
that gives the value 2.9055 K oms and this resistance is in series with 1.5 K ohms and 1.2 K ohms that gives the value 5.6055 K ohms now this resistance is in parallel with 5.6 K ohms resistance so the parallel combination gives
5.6055 X 5.6 5.6055 + 5.6
that gives the total resistance of the circuit with respect to source is 2.801 K ohms

Now for current go from bottom to top. For 18 V voltage source here there are two parallel resistances 5.6K and 5.6055K ohms now calculate current through 5.6055 ohms resistance that gives
18 5.6055
3.211 mA and this current is divided in 8.2 K ohms and 4.5 K ohms resistance as they are in parallel by current division rule the current in 4.5 K ohms combination is
3.211 x 8.2 8.2 + 4.5
that is 2.073 mA the same current will flow through 1.8 K ohms resistance as 4.5 K is the series combination of 1.8 K and 2.7 K ohms now from the given circuit to find I1 use KCL for the node between 5.6 K ohms resistor and 1.8 K ohms resistor for that we need current through 5.6 ohms resistor that is
18 56
3.214 mA i.e current through 5.6 K ohms resistor is 3.214 mA so we know the current through 1.8 K and 5.6 K resistors to find I1 we have to use KCL at the node between them i.e sum of currents entering the node is equal to sum of the currents leaving the node here both current through 5.6 K and 1.8 K are entering the node and current I1 is leaving the node so I1= current through 5.6K ohms + current through 1.8 K ohms=3.214+2.073 =5.287 mA . I1=5.287mA

voltage Vb means voltage at point B with respect to ground . b is located between 1.8K ohms and 2.7 K ohms so Vb can be found by knowing the voltage drops across resistance 2.7 K and 1.5 K ohms as the ground is present next to 1.5 K ohms the sum of voltage drops across 1.5 K ohms and 2.7K ohms gives the voltage at b with respect to ground as we know current through 2.7 K ohms is same as the current flowing in 1.8 K ohms as they are in series i.e.2.073 mA and the current through 1.5 K ohms resistor is 3.211 mA this is the same current flowing in 5.6055 K ohms combination

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Both the current are going up wards so by KVL Vb=(2.7$\times$(-2.073))+(1.5$\times$(-3.211))=-10.415V

for Vab as we know the value of Vb now calculate the value of Va in similar way and calculate Vab = Va-Vb. Va is the voltage across 1.2 K ohms i.e.1.2 K ohms is the only resistance between node a and the neutral or ground. current through resistance 1.2 K ohms is same as the current through 1.5 K i.e 3.211 mA so the voltage Va=(1.2$\times$3.211)=3.8532 V . Now as we know Va and Vb Vab=Va-Vb=3.853-(-10.415)=3.853+10.415 =14.27 V

Vab=14.27 V

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5.6055 X 5.6 5.6055 + 5.6

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3.211 x 8.2 8.2 + 4.5

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