In the circuit shown in figure. `epsi_(1)=3,epsi_(2)=2,epsi_(3)=6V,R_(1)=2,R_(4)=6Omega,R_(3)=2,R_(2)=4Omega and C=5muF`. Find the current in the resistor `R_(3)` and the electrical energy stored in the capacitor C.

In the circuit shown in figure , (epsilon)_(1)=3V,(epsilon)_(2)=2V,(epsilon)_(3)=1V and r_(1)=r_(2)=r_(3)=1(Omega) .Find the potential difference between the points A and B and the current through each branch.

In the given circuit E_1 = 3E_2 = 2E_3 = 6 volts R_1 = 2R_4 = 6ohms R_3 = 2R_2 = 4ohms C = 5 muf. Find the current in R and the energy stored in the capacitor.

In the circuit as shown in figure, E_(1)=2V , E_(2)=2V, E_(3)=1V, and R=r_(1)=r_(2)=r_(3)Omega . The potential difference between points A and B will be

In the circuit shown in Fig. 4.19. R_(1)=100 Omega, R_(2)=R_(3)=50 Omega, R_(4)=75 Omega and E=4.75 V. Work out the equivalent resistance of the circuit and the current in each resistor.

Find the currents going through the three resistors R_(v),R_(2)andR_(3) in the circuit of figure.

In the given circuit values are as follows epsi=2V,epsi_(2)=4V,R_(1)=1Omega and R_(2)=R_(3)=1Omega . Calculate the current through R_(1),R_(2) and R_(3)

The emf and resistances in the circuit of Fig 27-8a have the following values: epsi_(1)=4.4 V. epsi_(2)=2.1V , r_(1)=2.3 Omega, r_(2)=1.8 Omega, R=5.5 Omega (a) What is the current I in the circuit? (b) What is the potential difference between the terminals of battery 1 in Fig. 27 8a?

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