In the circuit shown the value of l in ampere is (B) 0.60 (D) 1.5 A40 (C) 0.4 4V 1.60 We - -

In Figure, if l_1 || l_2, the value of x is (a) 22 1/2 (b) 30 (c) 45 (d) 60

In triangleABC , Delta=6 , abc=60 , r=1 Then the value of 1/a+1/b+1/c is nearly (a) 0.5 (b) 0.6 (c) 0.4 (d) 0.8

In triangleABC , Delta=6 , abc=60 , r=1 Then the value of 1/a+a/b+a/c is nearly (a) 0.5 (b) 0.6 (c) 0.4 (d) 0.8

In Figure,if l_(1)l_(2), the value of x is 22(1)/(2) (b) 30 (c) 45 (d) 60

In the circuit shown in Fig. the emf of the sources is equal to xi = 5.0 V and the resistances are equal to R_(1) = 4.0 Omega and R_(2) = 6.0 Omega . The internal resistance of the source equals R = 1.10 Omega . Find the currents flowing through the resistances R_(1) and R_(2) .

In the circuit shown in Fig. the emf of the sources is equal to xi = 5.0 V and the resistances are equal to R_(1) = 4.0 Omega and R_(2) = 6.0 Omega . The internal resistance of the source equals R = 1.10 Omega . Find the currents flowing through the resistances R_(1) and R_(2) .

In the circuit shown below E_(1) = 4.0 V, R_(1) = 2 Omega, E_(2) = 6.0 V, R_(2) = 4 Omega and R_(3) = 2 Omega . The current I_(1) is

In the circuit shown below E_(1) = 4.0 V, R_(1) = 2 Omega, E_(2) = 6.0 V, R_(2) = 4 Omega and R_(3) = 2 Omega . The current I_(1) is