Two resistors of resistance 4 Ω and 6 Ω are connected in parallel to a cell to draw current 0.5 A from the cell.(a) Draw a labelled diagram of the arrangement.(b) Calculate the current in each resistor.

(a) As a text-based AI, I'm unable to draw diagrams. However, I can describe it for you. In the diagram, you would have a cell (or battery) on one side. Two lines (representing wires) would come out from the cell. These two lines would then connect to two separate resistors (one 4 Ω and the other 6 Ω), which are drawn as small rectangles. The other ends of the resistors would then connect back to the other terminal of the cell, completing the circuit. This represents a parallel circuit.

(b) In a parallel circuit, the voltage across each component is the same. The total current supplied by the cell is split between the components in the ratio of their resistances.

The total resistance (Rt) in a parallel circuit is given by the formula:

1/Rt = 1/R1 + 1/R2

Where R1 and R2 are the resistances of the individual resistors.

So, 1/Rt = 1/4 + 1/6 = 0.4167

Therefore, Rt = 1 / 0.4167 = 2.4 Ω

The total current (It) supplied by the cell is 0.5 A.

The voltage (V) across the cell can be calculated using Ohm's law (V = I * R):

V = It * Rt = 0.5 A * 2.4 Ω = 1.2 V

Now, we can calculate the current across each resistor using Ohm's law:

For the 4 Ω resistor: I = V / R = 1.2 V

(a) As a text-based AI, I'm unable to draw diagrams. However, I can describe it for you. In the diagram, you would have a cell (or battery) on one side. Two lines (representing wires) would come out from the cell. These two wires would then connect to the resistors, which are drawn as small rectangles. One resistor would have "4 Ω" written inside it and the other would have "6 Ω". The wires from both resistors would then connect back to the other end of the cell, completing the circuit.

(b) When resistors are connected in parallel, the total resistance (Rt) of the circuit can be calculated using the formula 1/Rt = 1/R1 + 1/R2. Here, R1 is 4 Ω and R2 is 6 Ω.

So, 1/Rt = 1/4 + 1/6 = 0.25 + 0.1667 = 0.4167. Therefore, Rt = 1 / 0.4167 = 2.4 Ω.

The total current (It) drawn from the cell is given as 0.5 A. According to Ohm's law (I = V/R), the voltage across the cell (V) can be calculated as V = It * Rt = 0.5 A * 2.4 Ω = 1.2 V.

Now, the current across each resistor (I1 for R1 and I2 for R2) can be calculated using Ohm's law again:

I1 = V / R1 = 1.2 V / 4 Ω = 0.3 A, I2 = V / R2 = 1.2 V / 6 Ω = 0.2 A.

So, the current in the 4 Ω resistor is 0.3 A and the current in the 6 Ω resistor is 0.2 A.