Two 27 2 kohm resistors are in series and this series combination is in parallel with a 29 3 kohm

***In a previous solution i had written (r1 + r2)/(r1 * r2) instead of
(r1 * r2)/(r1 + r2) for parallel combination ..... Now I have corrected that part in this solution.
two 27.2 kohm resistors are in series:
then total resistance in series combination is (27.2+27.2)kohm = 54.4 kohm
now 54.4 kohm and 29.3 kohm resistors are in parallel combination:
let r1 = 54.4 kohm and r2 = 29.3 kohm
total resistence is:
(r1 * r2)/(r1 + r2) = (54.4 * 29.3)/(54.4 + 29. 3)= 19.0432 Kohm
hence Total resistance is 19.0432 Kohm
11 years agoHelpfull: Yes(3) No(0)
19.04 k ohms

Final resistanmce = (27.2+27.2)*29.3/(27.2+27.2+29.3)=19.04 kohms
11 years agoHelpfull: Yes(2) No(0)
in series eqvlnt resistor=27.2*2=55.now eqvlnt resistor=(55*29.3)/(55+29.3)=19.116
11 years agoHelpfull: Yes(2) No(0)
Let R1=R2=27.2 Kohm connected in series and this series combination connected in parallel with a R3=29.3 Kohm resister.
therefore,equivalent resistance of series combination=R1+R2=(27.2+27.2)=54.4kohm
and finally total resistance of the circuit=(54.4*R3)/(54.4+R3)=19.04kohm
11 years agoHelpfull: Yes(1) No(0)

Two 27.2 kohm resistors are in series and this series combination is in parallel with a 29.3 kohm
resistor. The total resistance is