A series circuit consists of three identical lamps connected to a battery as shown in Fig. `7.13`.
when the switch `S` is closed, what happens
(a) to the intensities of lamps `A and B`,
(b) to the intensity of lamp `C`,
( c ) to the current in the circuit , and
(d) to the voltage drop across the three lamps ?
Does the power dissipated in the circuit increase, decrease, or remain the same ?

Explanation

As the three bulbs are in series and identical , initially when switch `S` is open,
`V_(A) = V_(B) = V_(C ) = (1)/(3) V`
Also `P_(A) = P_(B) = P_(C ) = ((V//3)^(2))/( R ) = ( V^(2)) /( 9 R) = P`
Now , the swtch `S` is closed,
(a) The bulb `C` is short - circuited and hence potential difference across it is `V' _(c ) = 0` and so `V'_(A) = V'_(B) = V/2` with `V'_(C)c = 0`. And hence,
`P'_(A) = P'_(B) = ((V//2)^(2))/( R ) = (V^(2))/( 4 R)`
i.e, `P'_(A) = P'(B) = ( 9)/(4) P`,
i.e, `P'_(A) = P'_(B) = (9)/(4) P`,ltbegt i.e, intensities of bulbs `A and B` will increase and become `2.25` of their initial values.
(b) As ` V'_( c ) = 0`, no current will pass through bulb ` C ` , so it will give no light , i.e, `P'_(c ) = 0 `.
As `R_(T)` changes from `3R to 2 R` , so that current in the circuit will change from
`I = ( V)/( 3 R) to I' = (V) / (2 R) , i.e, I' = (3) /(2) I`
Thus , current in the circuit will increase and will become `1.5` times its initial value.
(d) As explained earlier , initially the voltage `V` divides equally across `A, B, and C`, i.e, `V_(A) = V_(B) = V_(C) = V//3`. Now when the switch `S` is closed , bulb `C` is short - circuited , i.e, `V' _(C) = 0` , so voltage `V` now will divide equally across `A and B`, i.e, `V'_(A) = V'_(B) = V//2`. So voltage across bulbs `A and B` will change from `V//3 to V//2`, while across `C` from `V//3` to zero, i.e, voltage drop across `A and B` will increase while across `C` it will decrease and becomes zero. Further as `R_(T)` changes from `3 R to 2R`, so power consumed in the circuit changes from
`P_(T) = (V^(2))/(3 R ) to P'_(T) = (V^(2))/( 2 R ) , i.e, P'_(T) = (3)/(2) P_(T)`
Thus, power dissipated in the circuit increase and becomes `1.5` times its initial value.