Question

Consider the three identical light bulbs shown in the circuit. Bulbs B and C are wired in series with each other and are wired in parallel with bulb A. When the bulbs are connected to the battery as shown, how does the brightness of each bulb compare to the others?

a) Bulbs B and C are equally bright,

b) Bulbs B and C are equally bright, but less bright than bulb A.

c) All three bulbs are equally bright.

d) Bulbs A and B are bright, but bulb C is less bright.

e) Only bulb A is illuminated.​

Answer 1

Answer:

B is the answer.

Explanation:

correct answer

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Answer 2

Given that,

Bulbs B and C are wired in series with each other and are wired in parallel with bulb A.

Bulb A is wired in parallel with bulb B and this combination is wired in series with bulb C with equal resistance r.

Bulb A connected in parallel with B

We need to calculate the total resistance

Using formula of parallel

$\dfrac{1}{R}=\dfrac{1}{r}+\dfrac{1}{r}$

$R=\dfrac{r^}{2r}$  

Now, R and C in series

We need to calculate the effective resistance

Using formula of series

$R'=R+r$

Put the value of R into the formula

$R'=\dfrac{r^}{2r}+r$

$R'=\dfrac{3r}{2}$

We need to current in C

Using Ohm's law

$I=\dfrac{v}{R'}$

Put the value into the formula

$I=\dfrac{2v}{3r}$

Here, v = voltage

We need to calculate the current in A and B

Using ohm's law

$I=\dfrac{2v}{2\times3r}$

$I'=\dfrac{v}{3r}$

When Bulb B and C in series then the resistance will be 2r

So, The current in B and C will be

$I=\dfrac{v}{2r}$

We know that,

The brightness depends on the power of bulb.

The power is proportional to the current.

So, The power is greater in B and C compare to A.

Hence, Bulbs B and C are equally bright, but less bright than bulb A.

(b) is correct option.