The brightness of the bulbs is determined by the amount of the electric current flowing through them. So, in order to decide what will happen to the brightness of the bulbs B and C if bulb A burns out, we need to consider what happens to the electrical current in the circuit.

Since the combination of the bulbs A and B is wired in series with C, the total current in the original circuit is equal to the current C and total current in AB part:

`I = I_C = I_ (AB)`

To find this current, consider the total resistance of the original circuit. Since the bulbs A and B are wired in parallel, letting the resistance of each bulb to be R,

`1/R_(AB) = 1/R + 1/R`

From here, `R_(AB) = R/2` .

Since bulb C is wired in series with AB combination, the equivalent resistance of the circuit is `R_(circuit) = R + R/2 = 3R/2` .

Then the total current I, according to Ohm's Law, is

`I = U/((3R)/2) = (2U)/(3R)` , where U is the voltage supplied by the battery.

The current through C is then the same: `I_C = (2U)/(3R)` and this current is divided evenly between the A and B branches, since they are identical and connected in parallel:

`I_A = I_B = U/(3R)` .

Now let us see what happens if bulb A burns out. If it burns out, the current cannot flow through it and the A branch of parallel connection is no longer there. Now the circuit is just B and C connected in series. The new equivalent resistance is R + R = 2R.

Then the new total current is `I = U/(2R)` . Same current flows through B and C: `I_B = I_C = U/(2R)` .

Compare this expression with original currents through B and C:

`I_B` used to be `U/(3R)` and now is `U/(2R)` , so it increased, which means bulb B is brighter.

`I_C ` used to be `(2U)/(3R)` and now is `U/(2R)` , so it decreased, which means bulb C is dimmer.

**Choice 3 is correct.**

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