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, [tex]I_T[/tex] = 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;
[tex]I_2 = \dfrac{R}{R + R} \times I_T = \dfrac{R}{2 \cdot R} \times I_T = \dfrac{1}{2} \times I_T[/tex]
∴ 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 = [tex]I_T[/tex]