Three identical lamps each having a resistance R are connected to the battery of emf as shown in the figure.

Suddenly the switch S is closed. (a) Calculate the current in the circuit when S is open and closed (b) What happens to the intensities of the bulbs A,B and C. (c ) Calculate the voltage across the three bulbs when S is open and closed (d) Calculate the power delivered to the circuit when S is opened and closed (e ) Does the power delivered to the circuit decreases, increases or remain same?

Explanation

Resistance of the identical lamp = R
Emf of the battery `=xi`
According to Ohm.s Law, `xi=IR`
(a) Current: When Switch is open - The current in the circuit. Total resistance of the bulb,
`R_(s)=R_(1)+R_(2)+R_(3)`
`R_(1)=R_(2)=R_(3)=R`
`R_(s)=R+R+R=3R`
`therefore ` Current, `I=(x)/(R_(s)) rArr I_(0)=(xi)/(3R)`
Switch is closed - The current in the circuit. Total resistance of the bulb,
`R_(s)=R+R=2R`
`therefore ` Current, `I=(xi)/(R_(s)) rArr I_(c)=(xi)/(2R)`
(b) Intensity: When switch is open - All the bulbs glow with equal intensity.
When switch is closed - The intensities of the bulbs A and B equally increase. Bulb C will not glow since no current pass through it.
(c ) Voltage across three bulbs:
When switch is open - Voltage across bulb `A, V_(A)=I_(0)R=(xi)/(3R)xxR=(xi)/(3)`
Similarly: Voltage across bulb b, `V_(B)=(xi)/(3)`
Voltage across bulb C, `V_(C )=(xi)/(3)`
When switch is closed - Voltage across bulb A, `V_(A)=I_(c )R= (xi)/(2R)xxR=(xi)/(2)`
Similarly: Voltage across bulb B, `V_(B)=I_(c )R=(xi)/(2)`
Voltage across bulb C, `V_(C)=0`
(d) Power delivered to the circuit,
When switch is opened - Power P, `=VI`
`P_(A)=V_(A)I_(0)=(xi)/(3)xx(xi)/(3R)=(xi^(2))/(9R)`
Similarly: `P_(B)=(xi^(2))/(9R) and P_(c)=(xi^(2))/(9R)`
When switch is closed - Power P, `=VI`
`P_(A)=V_(A)I_(c)=(xi)/(2)xx(xi)/(2R)=(xi^(2))/(4R)`
Similarly:
`P_(B)=(xi^(2))/(4R) and P_(c)=0`
(e) Total power delivered to circuit increases.