The conbined resistance R of two resistors R,& `R_(2)(R_(1),R_(2) gt 0)` is given by . `(1)/(R)=(1)/(R_(1)) +(1)/(R_(2)).` If `R_(1)+R_(2)=` constant Prove that the maximum resistance R is obtained by choosing `R_(1)=R_(2)`

Calculate equivalent resistance of two resistors R_(1) and R_(2) in parallel where, R_(1) = (6+-0.2) ohm and R_2 = (3+-0.1)ohm

The capacitance of two concentric spherical shells of radii R_(1) and R_(2) (R_(2) gt R_(1)) is

Ketones (R_(1)COR_(2)):R_(1)=R_(2) =alkyl group, can be obtained in one step by

Two sources of equal emf are connected to an external resistance R. The internal resistance of the two sources are R_(1) and R_(2) (R_(1) gt R_(1)) . If the potential difference across the source having internal resistance R_(2) is zero, then

In the situation shown, resistance R_(1),R_(2) and R_(3) are in the ratio 3:2:1 , then

(r_(2)+r_(3))sqrt((r r_(1))/(r_(2)r_(3)))=

Prove the questions a(r r_(1) + r_(2)r_(3)) = b(r r_(2) + r_(3) r_(1)) = c (r r_(3) + r_(1) r_(2))

When resistors R_(1) and R_(2) are connected in a parallel electric circuit, the total resistance is (1)/((1)/(R_(1))+(1)/(R_(2))) . This fraction is equivalent to

Value of 1/(r_(1)^2)'+ 1/(r_(2)^2)+ 1/(r_(3)^2)+ 1/(r_()^2) is :

If in a triangle, (1-(r_(1))/(r_(2))) (1 - (r_(1))/(r_(3))) = 2 , then the triangle is