Question

Bulb A rated 160 W, 40 V and Bulb B rated 40 W, 40 V are connected as shown in the diagram.

1. Calculate the ratio V₁ : V₂.
2. If the bulb fuses, the current in the circuit remains the same.
   State True or False.

Answer 1

Given: Power rating of the bulb A (P_A​) = 160 W

Voltage rating of the bulb A (V_A​) = 40 V

Power rating of the bulb B (P_B​) = 40 W

Voltage rating of the bulb B (V_B) = 40 V

Supply voltage (V) = 40 V

a. Resistance of bulb A is given by,

R_A ​= `(V_A^2)/P_A`

= `40^2/160`

= `1600/160`

= 10 Ω

Similarly,

The resistance of bulb B is given by,

R_B ​= `V_B^2/P_B`

= `40^2/40`

= `1600/40`

= 40 Ω

From Ohm’s law,

Potential drop (V) = Current (I) × Resistance (R)

Since the bulbs are in series, the same current flows through both bulbs, and the potential difference divides in the ratio of their resistances, i.e.,

V₁ : V₂ = R_A : R_B

= 10 : 40

= 1 : 4

Hence, the ratio V₁ : V₂ = 1 : 4.

b. This statement is false.

Explanation:

If bulb A fuses, the circuit becomes open, so no current can flow through the circuit. Hence, the current becomes zero, not the same.