30 volt and 90 watt lamp is to be operated on a 120 volt dc land for proper bhumro how much resistance should be connected in series with the lamp
Answers
30 Ω resistance can be connected in the series.
Explanation:
Power rating of the bulb is, P = 90 W
Voltage rating, V = 30 V
Max. current the bulb can draw is, I = P/V = 90/30 = 3 A
So, resistance of the bulb is, Rb = V/I = 30/3 = 10 Ω
Let, R be the resistance connected in series to operate the bulb in 120 V supply.
So, (R + Rb)I = 120
=> R + 10 = 120/3 = 40
=> R = 30 Ω
Hence 30 Ω resistance can be connected in the series.
Also learn more
A 50 volt AC is applied across an RC (series) network. The rms voltage across the resistance is 40 V, then the potential across the capacitance will be
(A) 30 V
(B) 20 V
(C) 40 V
(D) 10 V
https://brainly.in/question/12178053
Answer:
30ohms
Explanation:
first we find the resistance of the lamp using the formula Power is equal to to (V square )/R = 10 ohms.
For proper glow the applied voltage voltage has to be brought to the voltage rating of the bulb.
So 120 volt has to be applied as 30 volt across the bulb and remaining 90 volt across the resistor.
Sins series connection the current is same across both the components.
So V/R is constant.
30/10=90/R
Therefore R is equal to 30 ohms.
Hope it helps you!!!!