- There is no circuit element between C and E. C and E are the same points.
- There is no circuit element between I and J. I and J are the same points.
- The simplified circuit can be plotted as follows.
- Separate the CDF triangle and apply a delta-star transformation.
- Now we have new resistances (R7, R8, R9)
- Let's connect the same points to the star circuit.
- Let us calculate
#R_7,R_8,R_9#
#R_7=(R_2*R_3)/(R_2+R_3+R_4)=(10*15)/(10+15+16)=150/41=3.66 Omega#
#R_8=(R_2*R_4)/(R_2+R_3+R_4)=(10*16)/(10+15+16)=160/41=3.90 Omega#
#R_9=(R_3*R_4)/(R_2+R_3+R_4)=(15*16)/(10+15+16)=240/41=5.85 OmegaOmega#
- Now that the circuit is simplified, we can easily solve it.
#R_1" and "R_7 " are connected in series."#
#R_9=R_1+R_7=20+3.66=23.66 Omega#
#R_9" and "R_6 " are connected in series."#
#R_10=R_9+R_6=10+5.85=15.85 Omega#
#R_8" and "R_5 " are connected in series."#
#R_11=R_8+R_5=12+3.90=15.90 Omega#
#R_10" and " R_11 " are connected in parallel."#
#R_12=(R_10*R_11)/(R_10+R_11)=(15.85*15.90)/(15.85+15.90)=7.94 Omega#
#R_9" and "R_12 " are connected in series."#
#Sigma R=R_9+R_12"(This is equivalent to resistance.)"#
#Sigma R=23.66+7.94=31.6 Omega#
#I=V/(Sigma R)=12/(31.6)=0.38 A#
#V_("AR")=I*R_9=0.38*23.66=8.99 V#
#V_("R J")=12-8.99=3.01V#
#I_6=(V_("R J"))/(R_("10"))=(3.01)/(15.85)=0.19A#
