Three `60 W , 120 V` light bulbs are connected across a `120 V` power lines as shown in Fig. `7.26`. Find (a) the voltage across each bulb and (b) the total power dissipated in the three bulbs.

Answer

(a) `80 V`, `40V`
(b) `40 W`

Explanation

(a) As bulbs `B and C ` are in parallel, voltage across `B and C` will be the same , i.e, `V_(B) = V_(C )` . Further , if `R` is the resistance of each bulb ( as bulbs are identical) , the resistance of bulbs `B and C` together `( = R//2)` is in series with resistance `R` of bulb `A` and as in series, potential divides in proportion to resistance.
`V_(A) = (R ) / ( R - 0.5 R) = ( 2) /(3) V_(s) = (2) /(3) xx (120) = 80 V`
`V _(B) = V_(C) = ( 0.5 R) /( R + 0.5 R) , V = (1)/( 3) V_(s) = (1) /(3) xx 120 = 40 V`
`[ or V_(B) = V_( C ) = V - V _(A) = 120 - 80 = 40 V]`
As actual power consumed by a bulb is
`P = (V_(A)^(2))/( R ) = [( V_(A))/( V_(S))]^(2) xx W` `[ because R = (V_(s)^(2))/( W)]`
So the total power consumed consumption is
` P = P_(A) + P_(B) + P_( C ) = P_(A) + 2P_(B) = ((4+2) 60)/(9) = 40 W `