The equivalent resistance between A - and B of the network shown in figure is
(a)`3R Omega`
(b) `((3)/(2))R Omega`
(c)`2R Omega`
(d)`((2)/(3))R Omega`

The equivalent resistance between A and B will be (in Omega )

The equivalent resistance between the point P and Q in the network given here is equal to (given r = (3)/(2) Omega)

(a) Find the equivalent resistance between A and B of the network extending off to the infinity shown in the figure . (b) If E = 12 V , R_(1) = 1 Omega and R_(2) = 4 Omega . Find the current in R_(2) nearest of the battery .

Find the current through the resistance R in figure.If (a) R=12 Omega (b) R=48 Omega .

A voltage V_(PQ)=V_(0) cos omega t , where V_(0) is a real amplitude, is applied between the points P and Q in the network shown in figure. Given that C=(1)/(omega R sqrt(3)) and L=(Rsqrt(3))/(omega) Calculate the total impedance between P and Q.

The potential difference in volt across the resistance R_(3) in the circuit shown in figure, is (R_(1)=15Omega,R_(2)=15Omega, R_(3)=30Omega, R_(4)=35Omega)

In the circuit shown in figure, the equivalent resistance between A and B is __________ Omega .

Current through resistor R3 as shown in figure is ______. Given that R1=10 Omega R2=5 Omega R4=20 Omega and R5=10 Omega

In the network shown in the above figure find the current through 3 Omega and 10 Omega resistances.

For infinite ladder network containing identical resistance of R Omega each, are combined as shown in figure. The equivalent resistance between A and B is R_(AB) and between A and C is R_(AC) . Then the value (R_(AB))/(R_(AC)) is :