Complete the table using the given three resistors and a power supply below with their respective resistances and voltage as:

R1 = 3Ω

R2 = 9 Ω

R3 = 18 Ω

ξ = 12 Volts

Community Answer

Step 1Given:-R1 =3 ohm R2 = 9 ohm R3 = 18 ohm V=12 VoltFind:-All the values of voltage and current for both circuit shown in the above Series - parallel combination parallel - series combination Step 2Solution:-a)For The Series Parallel combination:-We see that the Resistance  R2 and R3 are in parallel so the equivalent of them is {:[(1)/(R_(cq))=(1)/(R_(2))+(1)/(R_(3))=(1)/(9)+(1)/(18)],[R_(cq)=6ohm]:}R_(cq) resistance is series with the R_(1) so then{:[R_("net ")=R_(eq)+R_(I)=6+3=9ohm],[R_("net ")=9ohm]:}so The current in the circuit with V=12 volt{:[I=(12)/(9)=1.33A],[I_(T)=1.33A]:}So I_(1)=I=1.33A{:[I_(1)=1.33A],[V_(1)=1.33 xxR_(1)=1.33 xx3=3.99=4V],[V_(1)=4V]:}The R_(2) and R_(3) are in parallel so the voltage of R_(2) and R_(3) is same so V^(')=V_(2)=V_(3)and now We know that{:[V=V_(1)+V^(')=4+V^(')=12],[V^(')=12-4=8V]:}Step 3So the voltage of V_(2)=V_(3)=8Vso the current in R_(2) is I_(2) ... See the full answer