9 V की किसी बैटरी को 2Omega, 3Omega, 4Omega, 5 Omega तथा 12 Omega के प्रतिरोधकों के साथ श्रेणी क...

A battery of 9 v is connected in series with resistors of 0.2 Omega, 0.3 Omega, 0.4 Omega, 0.5 Omega and 12 Omega . How much current would flow through the 12 Omega resistor ?

In the figure shown, what is the current (in Ampere) drawn from the battery? You are given: R _ 1 = 15 Omega, R _ 2 = 10 Omega, R _ 3 = 20 Omega, R _4 = 5 Omega, R _ 5 = 25 Omega , R _ 6 = 30 Omega, E = 15 V

In figure. P=3Omega,Q=2Omega,R=6Omega,S=4Omega and X=5Omega . Calculate the current I.

(1-omega) (1-omega ^ (2)) (1-omega ^ (4)) (1-omega ^ (5)) = 9

In R_1 = 10 Omega, R_2 = 40 Omega and R_3 = 30 Omega, R_4 = 20 Omega, R_5 = 60 Omega and a 12 V battery is connected to the arrangement, calculate. (a) the total resistance and (b) the total current flowing in the circuit. .

(1-omega+omega^(2))(1+omega-omega^(2))=4