Two light bulbs are connected in series. The resistance of the filament of the first light bulb is 230Ω. The potential difference across the first light bulb is 100V and across the second light bulb,200V. What is the resistance of the second light bulb?
twice the voltage drop means twice the resistance
I = V1/R1 = 100/230 = 0.435A.
R2 = V2/I = 200/0.435 =
To find the resistance of the second light bulb, we need to apply the concept of voltage and resistance in a series circuit.
In a series circuit, the total resistance is equal to the sum of the individual resistances. Using Ohm's Law (V = IR), we can rearrange the formula to calculate the resistance: R = V/I.
Since the potential difference (V) across the first light bulb is 100V, and its resistance (R1) is known to be 230Ω, we can calculate the current (I1):
I1 = V1 / R1 = 100V / 230Ω = 0.4348A
Now, since the potential difference (V) across the second light bulb is given as 200V, we can use the current (I1) to calculate the resistance (R2) of the second light bulb:
R2 = V2 / I1 = 200V / 0.4348A = approximately 460.22Ω
Therefore, the resistance of the second light bulb is approximately 460.22Ω.