An energy source forces a constant current of 5 ampere for 4 sec to flow through a light bulb. If 3 kJ given in terms of light energy, the voltage drop across the bulb is________.

The correct answer is option 1):(150 V)

Concept:

The charge which flows given by

\(Δ q=iΔ t\)

where i is the current

Δt is time

The voltage drop across the bulb

V = \(Δw \overΔq ​ \)

Δw is the energy

Calculation:

i = 5 A

Δt = 4 sec

Δw = 3000 J

q = 5 × 4 =20 C

V = \(Δw \overΔq ​ \)

= \(3000 \over 20\)

= 150 V