Question

A student constructs four electric circuits.

Answer 1

Answer:

Part A

The ammeter reading for the current flowing through the second bulb, in the third circuit is 0.43 A

Part  B

The ammeter reading for the total current coming from the cell is 0.86 A

Explanation:

The given electric circuit schematics includes;

The number of bulbs in each circuit = 2 bulbs

The number of ammeter in each circuit = 1 ammeter

The number of cells in each circuit = 1

Each circuit have identical bulbs, cells and ammeters

Part A

Ammeter reading for the third circuit

The given reading for the total current going to the cell, $I_T$ = 0.86 A

The reading for the current flowing through one of the bulbs, I₁ = 0.43 A

Therefore, given that the bulbs are arranged in parallel, and the bulbs have identical resistance, 'R', the current flowing in the second bulb, I₂, is given by the current divider rule as follows;

$I_2 = \dfrac{R}{R + R} \times I_T = \dfrac{R}{2 \cdot R} \times I_T = \dfrac{1}{2} \times I_T$

∴ The ammeter reading for the current flowing through the second bulb, in the third circuit I₂ = (1/2) × (0.86 A) = 0.43 A

Part B

Ammeter reading for the third circuit

The total current coming from the cell = The current passing through the bulbs = I₁ + I₂

The ammeter reading for the total current coming from the cell = 0.43 A + 0.43 A = 0.86 A = $I_T$