A series circuit consists of three bulbs connected to a battery as shown. When switch `S` is closed, what happens to (a) Power consumed in bulb `X` and `Y` (b) Power consumed in bulb `Z` (c ) the current in the circuit (d) the voltage drop across three bulbs and (e) the power consumed in circuit?

Explanation

Let `R` : resistance of each bulb
When `S` is open
`V_(X) = V_(Y)=V_(Z)=V//3`
`P_(X) = P_(Y) = P_(Z)= (V^2)/(9R)`
Current in the circuit `I = V/(3R)`
Power consumed in the circuit `P = i^(2) xx 3R = (V^2)/(3R)`
When `S` is closed:

Current in bulb `Z` is zero
`V_(X)^(') = V_(Y)^(') = V/2, V_(Z)^(') = 0, P_(X)^(') = P_(Y)^(') = (V^2)/(4R) P_(Z)^(') = 0`
Current in the circuit `i' = V/(2R)`
Power consumed `P' = i_(2)^(') xx 2R = (V^2)/(2R)`
(a) `(P_(X)^(') = P_Y^') gt (P_(X)=P_(Y))`
(b) `P'_(Z) lt P_(Z)`
(c) `i' gt i`
(d) `(V'_(X) = V'_(Y)) gt V_X = V_Y) , V'_Z lt V_Z`
(e) `P' gt P`.