For three resistors connected in parallel,prove that the equivalent resistance is equal to the sum of the individual resistance Rp is given by`1/R_P=1/R_1+1/R_2+1/R_3`

Four resistance R_1,R_2,R_3 and R_4 are connected in parallel. The resultant resistance R is

Two resistance r_1 and r_2 (r1< r2) are joined in parallel. The equialent resistane R is such that

Three resistance R_1 , R_2 and R_3 are connected on parallel. Show that their equivalent resistance is 1/R=1/R_1+1/R_2+1/R_3

Three unequal resistor in parallel are equivalent to a resistance 1Omega If two of them are in the ratio 1:2 and if no resistance value is fractional the largest of the three resistance in ohm .

Resistance n, each of r ohm, when connected in parallel give an equivalent resistance of R ohm. If these resistances were connected series, the combination would have a resistance in ohm, equal to

n conductors, each of resistance r, when connected in parallel, give an effetive resistance R. What will be resistance, if they are connected in series?

When two resistors of resistance R_1 and R_2 are connected in parallel, the net resistance is 3 ohm. When connected in series. Its values of 16 ohm. Calculate the value of R_1 and R_2 .

With two resistance R_1 and R_2 (>R_1) in the two gaps of a metre bridge the balance was found to be 1/3 m from the zero end. When a 6 Omega resistance is connected in series with the smaller of the two resistance, the point is shifted to 2/3 m from the same end, then R_1 and R_2 Find the resistance of the two wires.