Question

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.

1. Calculate the current in the circuit when S is open and closed.
2. What happens to the intensities of the bulbs A, B and C.
3. Calculate the voltage across the three bulbs when S is open and closed.
4. Calculate the power delivered to the circuit when S is opened and closed.
5. Does the power delivered to the circuit decrease, increase or remain same?

Answer 1

Resistance of the identical lamp = R

Emf of the battery = ξ

According to Ohm’s Law, ξ = IR

(a) Current:

When Switch is open - The current in the circuit.

Total resistance of the bulb,

R_s = R₁ + R₂ + R₃

R₁ = R₂ = R₃ = R

R_s = R + R + R = 3R

∴ Current, I = `ξ/"R"_"s"`

`=> "I"_0 = ξ/(3"R")`

Switch is closed- The current in the circuit.

Total resistance of the bulb,

R_s = R + R = 2R

Current I = `ξ/"R"_"s"`

`"I"_"c" = ξ/(2"R")`

(b) Intensity:

When a switch is open - All the bulbs glow with equal intensity.

When a 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 = I₀ R = `ξ/"3R"`, R = `ξ/3`

similarly:

Voltage across bulb B, V_B = `ξ/3`

Voltage across bulb C, V_C = `ξ/3`

When switch is closed— Voltage across bulb A, V_A = I_cR = `ξ/"2R" xx ξ/2`

similarly:

Voltage across bulb B, V_B = I_cR `ξ/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 = ξ/3 xx ξ/"3R" = ξ^2/"9R"`

Similarly: `"P"_"B" = ξ^2/"9R" and "P"_"c" = ξ^2/"9R"`

When switch is closed — Power P, = VI

`"P"_"A" = "V"_"A""I"_"c" = ξ/2 xx ξ/"2R" = ξ^2/"4R"`

Similarly: `"P"_"B" = ξ^2/"4R" and "P"_"c" = 0`

(e) Total power delivered to circuit increases.