For the circuit shown in the diagram given: Calculate the total current drawn from the battery.

From the diagram, Potential difference V = 6 V , Let R 1 , R 2 , R 3 be the three resistors having the resistance value 12 Ω , 18 Ω , 30 Ω respectively. Using Ohm's law, we can find the total current flowing through the circuit. To find the total resistance, The resistors are connected in parallel combination, the formula is given by, 1 R P = 1 R 1 + 1 R 2 + 1 R 3 + . . . Substituting, 1 R P = 1 12 + 1 18 + 1 30 ⇒ 1 R P = 31 180 ⇒ R P = 180 31 ⇒ R P = 5 . 8 Ω I = V R = 6 5 . 8 = 1 . 03 A The total current drawn from the battery is 1 . 03 A .

For the circuit shown in the diagram given: Calculate the value of the current through each resistor.

Given, potential difference across the circuit is 6 V . Now Using Ohm's law: V = I R I = V R Since the circuit is connected in parallel, the potential drop across resistors remains same. Current through 12 Ω resistor = 6 12 = 0 . 5 A Current through 18 Ω resistor = 6 18 = 0 . 33 A Current through 30 Ω resistor = 6 30 = 0 . 2 A

In the given circuit diagram, calculate: Total resistance between the points A and F .

Here R 2 , R 3 and R 4 are in series. If its equivalent is R s then R s = R 2 + R 3 + R 4 = 1 + 2 + 3 = 6 Ω Now R s and R 5 are in parallel. If its equivalent is R p then R p = 6 × 3 6 + 3 = 2 Ω Now R 1 , R p and R 6 are in series. The equivalent resistance is R eq = R 1 + R p + R 6 = 3 + 2 + 3 = 8 Ω