
A 220 Volt, 1000 W bulb is connected across a 110 Volt mains supply. The power consumed will be:
A) 750 Watt
B) 500 Watt
C) 250 Watt
D) 1000 Watt
Answer
232.8k+ views
Hint: To solve this question use the formula of power and find the resistance of the bulb and you know the resistance of anything always remains constant. After that again by using the formula of power consumption we can simply calculate the power consumption by bulb when it connected to 110 Volt supply.
Complete step by step solution:
Step 1: First of all we have to calculate the resistance of the bulb by using information whatever is written on the bulb. In this question given the bulb is 220 Volt, and 1000 Watt.
As we know that the relation between power and voltage and resistance of a bulb is given as.
$P = \dfrac{{{V^2}}}{R}$ Where $P \Rightarrow $ power written on bulb means maximum power of bulb.
$V \Rightarrow $ Voltage written on the bulb on which it can operate.
$R \Rightarrow $ Resistance of bulb which is constant.
In question $P = 1000 W$, $V = 220V $ put these value in above equation
$ \Rightarrow 1000 = \dfrac{{{{220}^2}}}{R}$
$ \Rightarrow R = \dfrac{{220 \times 220}}{{1000}}$
Solving this
$ \Rightarrow R = \dfrac{{484}}{{10}}\Omega $
So here we got the resistance of bulb $R = \dfrac{{484}}{{10}}\Omega $
Step 2: As in question it is given when this bulb connected to $V = 110 V$ supply then
The power consumption by bulb $P$ can be given by
$ \Rightarrow P = \dfrac{{{V^2}}}{R}$
After putting value of $R$ and supply $V$
$ \Rightarrow P = \dfrac{{{{110}^2}}}{{\dfrac{{484}}{{10}}}}$
$ \Rightarrow P = \dfrac{{110 \times 110 \times 10}}{{484}}$
Solving it
$ \Rightarrow P = 250 Watt$
$\therefore $ When this bulb connects to 110 V supply then it will consume 250 Watt power.
Hence option (C) is correct.
Note: Sometimes students get confused. We use the same formula but put the different values. Remember when we apply the first time $P = \dfrac{{{V^2}}}{R}$ we find the resistance of the bulb which always remains the same and second time we use formula then we calculate the power of same bulb when it connected to 110 V instead of 220 V means if we connect this bulb with 220 V then it light with its full power but if it connected to 110 V then it light like a 250 watt power bulb .
Complete step by step solution:
Step 1: First of all we have to calculate the resistance of the bulb by using information whatever is written on the bulb. In this question given the bulb is 220 Volt, and 1000 Watt.
As we know that the relation between power and voltage and resistance of a bulb is given as.
$P = \dfrac{{{V^2}}}{R}$ Where $P \Rightarrow $ power written on bulb means maximum power of bulb.
$V \Rightarrow $ Voltage written on the bulb on which it can operate.
$R \Rightarrow $ Resistance of bulb which is constant.
In question $P = 1000 W$, $V = 220V $ put these value in above equation
$ \Rightarrow 1000 = \dfrac{{{{220}^2}}}{R}$
$ \Rightarrow R = \dfrac{{220 \times 220}}{{1000}}$
Solving this
$ \Rightarrow R = \dfrac{{484}}{{10}}\Omega $
So here we got the resistance of bulb $R = \dfrac{{484}}{{10}}\Omega $
Step 2: As in question it is given when this bulb connected to $V = 110 V$ supply then
The power consumption by bulb $P$ can be given by
$ \Rightarrow P = \dfrac{{{V^2}}}{R}$
After putting value of $R$ and supply $V$
$ \Rightarrow P = \dfrac{{{{110}^2}}}{{\dfrac{{484}}{{10}}}}$
$ \Rightarrow P = \dfrac{{110 \times 110 \times 10}}{{484}}$
Solving it
$ \Rightarrow P = 250 Watt$
$\therefore $ When this bulb connects to 110 V supply then it will consume 250 Watt power.
Hence option (C) is correct.
Note: Sometimes students get confused. We use the same formula but put the different values. Remember when we apply the first time $P = \dfrac{{{V^2}}}{R}$ we find the resistance of the bulb which always remains the same and second time we use formula then we calculate the power of same bulb when it connected to 110 V instead of 220 V means if we connect this bulb with 220 V then it light with its full power but if it connected to 110 V then it light like a 250 watt power bulb .
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