A student builds the circuit, as shown, using a battery of negligible internal resistance. The EMF of cell is 12 V . The voltmeter has a resistance of 12 kΩ . Show that the reading on this voltmeter would be 9 . 5 V .

The R e q u , of the system is given as 1 R 1 = 1 400 + 1 12000 = 31 12000 R 1 = 12000 31 R e q u = R 1 + 100 R e q u = 12000 31 + 100 R e q u = 15100 31 The current is given circuit I = V R e q u = 12 15100 31 = 372 15100 = 25 1000 Applying Kirchhoff's Voltage Law in given loop, V c - i R 100 - V 12 K = 0 12 - 2 . 5 = V 12 K V 12 K = 9 . 5 V

This diagram shows a network of resistors connected to a cell of e.m.f. 6 . 0 V . Show that the resistance of the network of resistors is 40 Ω .

The equivalent resistance R e q u , of the given circuit is given as, The parallel equivalent of the resistance is given as, 1 R = 1 R 1 + 1 R 2 The series equivalent of the resistance is given as R = R 1 + R 2 1 R 1 = 1 40 + 1 60 1 R 1 = 3 + 2 120 = 5 120 = 1 24 R 1 = 24 Ω R 2 = R 1 + 96 = 24 + 96 = 120 Ω R 3 = 40 + 20 = 60 Ω 1 R e q u = 1 R 2 + 1 R 3 1 R e q u = 1 120 + 1 60 1 R e q u = 1 + 2 120 = 3 120 = 1 40 R e q u = 40 Ω

This diagram shows a network of resistors connected to a cell of e.m.f. 6 . 0 V . Calculate the current in the 60 Ω resistor.

The equivalent resistance R e q u , of the given circuit is given as, The parallel equivalent of the resistance is given as, 1 R = 1 R 1 + 1 R 2 The series equivalent of the resistance is given as R = R 1 + R 2 1 R 1 = 1 40 + 1 60 1 R 1 = 3 + 2 120 = 5 120 = 1 24 R 1 = 24 Ω R 2 = R 1 + 96 = 24 + 96 = 120 Ω R 3 = 40 + 20 = 60 Ω 1 R e q u = 1 R 2 + 1 R 3 1 R e q u = 1 120 + 1 60 1 R e q u = 1 + 2 120 = 3 120 = 1 40 R e q u = 40 Ω The current in given circuit is I = V R = 6 40 3 20 A From the parallel resistor voltage. I 2 R 2 = I 1 R 3 I 2 R 2 = I - I 2 R 3 I 1 + I 2 = I I 2 = I R 3 R 2 + R 3 = 6 40 × 60 120 + 60 = 6 × 60 40 × 180 = 1 20 A 60 I 3 = 40 I 4 60 I 3 = 40 I 2 - I 3 I 3 + I 4 = I 2 100 I 3 = 40 I 2 I 3 = 40 100 × 1 20 = 1 50 A