Determine the voltage drop across the resistor `R_(1)` in the circuit given below with `E=60 V, R_(1)=18 Omega, R_(2)=10 Omega`.

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.

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)

If the voltage across 500 Omega is 5 V. then R_2 is

Find a potentail difference varphi_(1) - varphi_(2) between points 1 and 2 of the circuit shown in Fig if R_(1) 1= 10 Omega, R_(2) = 20 Omega, E_(1) = 5.0 V , and E_(2) = 2.0 V . The internal resisances of the current sources are negligible.

An alternating voltage V = 100 sin omega t is applied across an LCR circuit as shown. At the instant when voltage drop across source is 50 sqrt(3) volts then at that instant R = 30 Omega" "X_(L ) = 60 Omega" "X_(C ) = 20 Omega

(a) Resistors given as R_(1), R_(2) and R_(3) are connected in series to a battery V. Draw the circuit diagram showing the arrangement. Derive an expression for the equivalent resistance of the combination. (b) If R_(1)=10 Omega, R_(2)=20 Omega and R_(3)=30 Omega , calculate the effective resistance when they are connected in series to a battery of 6 V. Also find the current flowing in the circuit.

In the circuit shown, the voltage drop across the 15Omega resistor is 30V having the polarity as indicated. The ratio of potential difference across 5Omega resistor and resistance R is :

In the given cirucit, determine current through branch having indicated resistor R_(4) Given R_(1) = R_(2) = 4Omega, R_(3) = R_(4) R_(5) = 2Omega epsilon_(1) =5 V, epsilon _(2) =10V