Question

Three resistors, R1-2.0 Ω, R2-4.0 Ω, and R3 6.0 Ω are connected as shown in the figure below. For Circuit 1 (left), (a) Find the equivalent resistance of the combination; t.e. Req1] in Ohms ) Find the current that passes through the combination if a potential of 8.0 V is applied to the terminals; L.e. [itot1) in Amps. (c) Find the voltage across and the current through each resistor; L.e. R1: Voltage: [VR11] V and Current: [IR11] A R2: Voltage: [VR21] V and Current: [IR21] A R3: Voltage: [VR31] V and Current: [IR31) A

Answer 1

LEFT

R2 R3 parallel followed by R1 series

$R_{eq}=2+(4^{-1}+6^{-1})^{-1}=22/5=4.4\Omega$

Total current

$I=V/R_{eq}=8/4.4\approx1.82 \textup{ Amp}$

V across R2=V across R3=$IR_{23}=2.4*1.82=4.36\ V$

V across R1=$IR_{1}=2*1.82=3.63\ V$

I in R2

$I_{2}=\frac{R_3}{R_2+R_3}I_{Tot}=1.82*0.6=1.1\ Amp$

I in R3

$I_{3}=\frac{R_2}{R_2+R_3}I_{Tot}=1.82*0.4=0.72\ Amp$

I in R1 is same as the total current, i.e., 1.82 A

RIGHT

All 3 resistance are in parallel.

$R_{eq}=\left (2^{-1}+4^{-1}+6^{-1} \right )^{-1}\approx1.1\Omega$

Voltage across each is 8 V.

$I_1=V/R_1=4\textup{ Amp}$

$I_2=V/R_2=2\textup{ Amp}$

$I_3=V/R_3=1.33\textup{ Amp}$

Total current

$I_1+I_2+I_3$

or simply

$V/R_{eq}=8/1.1=7.33\textup{ Amp}$