
An electric bulb is marked $100\,W, 230\,V$. If the supply voltage drops to $115\,V$, what will be the total heat and light energy, totally liberated by the bulb in $20$ minutes?
Answer
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Hint:We will start answering this question, by first stating what power is, later we will use the formula of power in terms of current and voltage. After finding current we will next find power using the same formula and at last find energy.
Formula used:
\[P = V \times I\]
Where,$P$=power, $V$=voltage and $I$=current.
\[E = P \times T\]
Where $E$= energy, $P$= power and $t$= time.
Complete step by step answer:
In this question, the bulb is rated at 100 watts and 230 volts. The maximum current drawn by the bulb is calculated as follows: we will use the formula for power in terms of current and voltage. The link between the quantity of Electric Power (Watt) and the amount of Voltage (Volt) and Electric Current (Ampere) that flows is as follows:
\[P = V \times I\]
The higher the voltage (Volt), the bigger the power (Watt) The higher the current (ampere), the higher the power (Watt). Let's start solving.
\[P = V \times I\]
\[\Rightarrow I = \dfrac{P}{V}\]
\[\Rightarrow I = \dfrac{{100}}{{230}}\]
\[\Rightarrow I = \dfrac{{10}}{{23}}A\]
When the voltage is 115V, the power consumed can be calculated as follows:
\[P = V \times I\]
\[\Rightarrow P = 115 \times \dfrac{{10}}{{23}}\]
\[\Rightarrow P = \dfrac{{1150}}{{23}}W\]
\[\Rightarrow P = 50\,W\]
However, it is lit for 20 minutes. As a result, the total energy consumed is equal to
\[E = P \times T\]
\[\Rightarrow E = 50 \times 20 \times 60J\].................(converting minutes into seconds).
\[\therefore E = 60000\,J\].
Hence, the total heat and light energy, totally liberated by the bulb in 20 minutes if the supply voltage drops to 115V from 230V is 60000 Joules.
Note:Students can make a very common mistake in the second step while finding power: consider the voltage 150V and not 230V because we are finding power for the reduced voltage. Another very silly mistake is not writing the units. Please remember to write the units of every term that you answer.
Formula used:
\[P = V \times I\]
Where,$P$=power, $V$=voltage and $I$=current.
\[E = P \times T\]
Where $E$= energy, $P$= power and $t$= time.
Complete step by step answer:
In this question, the bulb is rated at 100 watts and 230 volts. The maximum current drawn by the bulb is calculated as follows: we will use the formula for power in terms of current and voltage. The link between the quantity of Electric Power (Watt) and the amount of Voltage (Volt) and Electric Current (Ampere) that flows is as follows:
\[P = V \times I\]
The higher the voltage (Volt), the bigger the power (Watt) The higher the current (ampere), the higher the power (Watt). Let's start solving.
\[P = V \times I\]
\[\Rightarrow I = \dfrac{P}{V}\]
\[\Rightarrow I = \dfrac{{100}}{{230}}\]
\[\Rightarrow I = \dfrac{{10}}{{23}}A\]
When the voltage is 115V, the power consumed can be calculated as follows:
\[P = V \times I\]
\[\Rightarrow P = 115 \times \dfrac{{10}}{{23}}\]
\[\Rightarrow P = \dfrac{{1150}}{{23}}W\]
\[\Rightarrow P = 50\,W\]
However, it is lit for 20 minutes. As a result, the total energy consumed is equal to
\[E = P \times T\]
\[\Rightarrow E = 50 \times 20 \times 60J\].................(converting minutes into seconds).
\[\therefore E = 60000\,J\].
Hence, the total heat and light energy, totally liberated by the bulb in 20 minutes if the supply voltage drops to 115V from 230V is 60000 Joules.
Note:Students can make a very common mistake in the second step while finding power: consider the voltage 150V and not 230V because we are finding power for the reduced voltage. Another very silly mistake is not writing the units. Please remember to write the units of every term that you answer.
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