Which bulb will glow brighter in series wiring?
In series connections, we can always encounter with a popular question, which bulb will glow brighter when the lamps have different power ratings.
We know that the current through a series path will be same at every point in the circuit. Only the voltage drops across the conductor path, which is in proportional to the resistance in the loop. That is based on a voltage division rule.
In the case of lamps with equal pure resistance (ideal resistive load), the voltage drops equally across all the lamps and they all glow with same brightness.
For a Lamps, with different power rating, the variation of brightness depends on the power dissipation across individual loads.
So considering in terms of the power dissipation the load with higher power dissipation have more brightness.
Power dissipation P= VI = I2R
V – voltage drop across the Load.
I – current.
R – resistance of the Load (Considering a purely resistive load).
In the case of Load L1 with 60W – 230V,50HZ and L2 with 100W- 230V, 50HZ. The power dissipation can be calculated as,
The Power dissipated on load L1, P1 = I2 *R1 & Power dissipated on load L2, P1 = I2 *R2
I2 is same for L1 and L2, then P1 and P2 is proportional to R1 and R2 respectively.
If we consider the Power dissipation in terms of voltage and current,
P1 = V1 * I, P2 =V2*I
Since the value of current through the both resistances are same, the bulb with high voltage drop will have the high power dissipation. That is the load with higher resistance dissipates more power. Because the voltage drop across a resistor is the product of current and resistance (V=IR).
For two loads which have different power rating and same rated operating voltage. The load resistance will be inverse to the power rating. That is higher the power rating, the load resistance will be less. Which means the amount of current required to be flowing for its rated power dissipation requires a higher value. Compared with the low resistance load, the bulb with higher resistance has less difference between its rated current and the actual current in the circuit. Thereby less difference in the rated power and actual power dissipation value. Hence, here the bulb L1 with low rated power and high resistance will glow brighter.
In the case of two loads with different power and voltage rating. We should have to find the power dissipation across the individual loads or the load with higher resistance value. We can’t simply say that the bulb with a low power rating will glow brighter. Because, the variation in operating voltage means that the value of the resistance also has variations, thereby the power dissipation.
P= p1+p2
=60w+100w
=160w
P=v^2/R
R=230^2/160
=330.625ohm=R1+R2
I=V/R=230÷330.625=.6956A
P1=I^2×R1
R1=P1÷I^2 =60÷(.6956)^2=124ohm
P2=I^2×R2
R2=P2÷I^2 =100÷(.6956)^2 =206.67ohm
V1=I R1=.6956×124=86.25v
V2=I R2=.6956×206.67=143.76v
V(total)=V1+V2=230v
I think the bulb consume 100w is brighter
Sorry ! my mistake
I think you got the point. Sometimes it is a little bit confusing.
To find the individual resistance of each bulb, the easy and simple way is calculating them separately with its rated voltage and power.
Hence the R1 = Vrated^2/P1 & R2 = Vrated^2/P2
P = p1+p2 =60w + 100w =160w, can be applied in a parallel connection. For a series connection, we need to take the power as reciprocal of the sum of the reciprocals, like a resistance in parallel.