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)`

Find the current flowing through the resistance R_(1) of the circuit shown in figure-3.103. the resistances are equal to R_(1)=10Omega,R_(2)=20Omega and R_(3)=30Omega and the potentials of points 1,2 and 3 are given as V_(1)=10V,V_(2)=6V and V_(3)=5V .

Determine the voltage drop across the resistor R_1 in the circuit given below with E = 65V, R_1 = 50 OmegaR_2 = 100 Omega, R_3 = 100 Omega, and R_4 300 Omega.

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)=20Omega,R_(4)=5Omega , R_(5)=25Omega,R_(6)=30Omega,E=15V

Passage : A battery E_(A) of 123 volts emf internal resistance 0.4Omega and a battery E_(B) of 117 V emf and internal resistance 0.6Omega supplies a current to load resistance R_(L) through four wires of resistors R_(A)=1.2Omega, R_(B)=1Omega, R_(C )=1.4Omega and R_(D)=1.4Omega as shown in figure. The value of load resistance for maximum load power is

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 circuit shown in figure Current through R_(2) is zero if R_(4) = 2 Omega and R_(3) = 4 Omega In this case

In the circuit shown in figure Current through R_(2) is zero if R_(4) = 2 Omega and R_(3) = 4 Omega In this case

In Fig. 27-43, the current in resistance 6 is i_(6)= 2.80 A the resistances are R_(1)=R_(2)=R_(3)=2.00 Omega, R_(4)=16.0 Omega, R_(5)=8.00 Omega and R_(6)=4.00 Omega . Wat is the emf of the ideal battery?