Complete Question
The complete question is shown in the first uploaded image
Answer:
a
When the both bulb are in the circuit bulb B glows equally brighter to bulb A
This because the power delivered to the both bulb are equal
b
The bulb A on the right will glow brighter than the bulb A on the left due to the fact that the power supplied to bulb A on the right is higher than that gotten by bulb A on the left.
Explanation:
From the question we are been told that the two bulbs are identical
So their resistance denoted by R is the same
Considering the left circuit where the two bulbs are connected in series which mean that the same current is passing through them
[tex]R_A =R_B =R[/tex]
[tex]i_A = i_B =i[/tex]
[tex]R_{eq} = R_1 +R_2 = 2R[/tex]
[tex]i = \frac{V}{2R}[/tex]
The power that is been deposited on the circuit is evaluated as
[tex]P_A = i^2R[/tex]
[tex]P_A = \frac{V^2}{4R}[/tex]
[tex]P_B = i^2R[/tex]
[tex]P_B = \frac{V^2}{4R}[/tex]
For the fact that the power deposited on the bulbs are the same they will glow equally
When B is now removed and only A is left
[tex]R_{eq} = R_A = R[/tex]
[tex]i = \frac{V}{R}[/tex]
[tex]P'_A = i^2R[/tex]
[tex]P'_A = \frac{V^2}{R}[/tex]
For the fact that its only bulb A that is on that right circuit the power delivered to it would be greater compared to the left circuit bulb A
