
Two lamps, one rated $100\,W$; $220\,V$, and the other $60\,W$; $220\,V$ are connected in parallel to the electric mains supply. Find the current drawn by two bulbs from the line, if the supply voltage is $220\,V$.
Answer
233.1k+ views
Hint: To find the resistance of the lamp, use the formula given below and substitute the known values in it. From the calculated values of the resistance and the supply voltage given, calculate the current flowing the lamp. Repeat the same process to find for the second lamp.
Useful formula:
(1) The resistance of the lamp is given by
$R = \dfrac{{{V^2}}}{P}$
Where $R$ is the resistance of the lamp, $V$ is the voltage of the lamps and $P$ is the power of the lamp.
(2) The ohm’s law is given as
$V = IR$
Where $I$ is the current required for the lamp.
Complete step by step solution:
It is given that the
The rating of the first lamp, $P = 100\,W$ and $V = 220\,V$
The rating of the second lamp, $p = 60\,W$ and $V = 220\,V$
The supply voltage, $SV = 220\,V$
By using the formula of the resistance,
$\Rightarrow$ $R = \dfrac{{{V^2}}}{P}$
By using the values in the above formula.
$\Rightarrow$ $R = \dfrac{{{{220}^2}}}{{100}}$
By simplifying the above step,
$\Rightarrow$ $R = \dfrac{{48400}}{{100}}$
By the further simplification,
$\Rightarrow$ $R = 484\,\Omega $
Using the formula (2),
$V = IR$
By rearranging the formula,
$\Rightarrow$ $I = \dfrac{{SV}}{R}$
By substituting the known values,
$\Rightarrow$ $I = \dfrac{{220}}{{484}}$
$\Rightarrow$ $I = 0.45\,A$
Similarly, calculating the resistance and the current of the second lamp.
$\Rightarrow$ $R = \dfrac{{{v^2}}}{p}$
Substituting the known values,
$\Rightarrow$ $R = \dfrac{{{{220}^2}}}{{60}}$
$\Rightarrow$ $R = 806.67\,\Omega $
By using the resistance of the lamp in the ohm’s law to find the value of the current.
$\Rightarrow$ $I = \dfrac{{SV}}{R}$
$\Rightarrow$ $I = \dfrac{{220}}{{806.67}}$
By further simplification,
$\Rightarrow$ $I = 0.27\,A$
Hence the current drawn by the first lamp is $0.45\,A$ and the current drawn by the second lamp is $0.27\,A$.
Note: Ohm’s law states that the voltage is directly proportional to the current. The first formula of the resistance is derived from the formula $P = VI$ . In this formula, $I = \dfrac{V}{R}$ is substituted as $P = \dfrac{{V \times V}}{R}$. It is rearranged as $R = \dfrac{{{V^2}}}{P}$.
Useful formula:
(1) The resistance of the lamp is given by
$R = \dfrac{{{V^2}}}{P}$
Where $R$ is the resistance of the lamp, $V$ is the voltage of the lamps and $P$ is the power of the lamp.
(2) The ohm’s law is given as
$V = IR$
Where $I$ is the current required for the lamp.
Complete step by step solution:
It is given that the
The rating of the first lamp, $P = 100\,W$ and $V = 220\,V$
The rating of the second lamp, $p = 60\,W$ and $V = 220\,V$
The supply voltage, $SV = 220\,V$
By using the formula of the resistance,
$\Rightarrow$ $R = \dfrac{{{V^2}}}{P}$
By using the values in the above formula.
$\Rightarrow$ $R = \dfrac{{{{220}^2}}}{{100}}$
By simplifying the above step,
$\Rightarrow$ $R = \dfrac{{48400}}{{100}}$
By the further simplification,
$\Rightarrow$ $R = 484\,\Omega $
Using the formula (2),
$V = IR$
By rearranging the formula,
$\Rightarrow$ $I = \dfrac{{SV}}{R}$
By substituting the known values,
$\Rightarrow$ $I = \dfrac{{220}}{{484}}$
$\Rightarrow$ $I = 0.45\,A$
Similarly, calculating the resistance and the current of the second lamp.
$\Rightarrow$ $R = \dfrac{{{v^2}}}{p}$
Substituting the known values,
$\Rightarrow$ $R = \dfrac{{{{220}^2}}}{{60}}$
$\Rightarrow$ $R = 806.67\,\Omega $
By using the resistance of the lamp in the ohm’s law to find the value of the current.
$\Rightarrow$ $I = \dfrac{{SV}}{R}$
$\Rightarrow$ $I = \dfrac{{220}}{{806.67}}$
By further simplification,
$\Rightarrow$ $I = 0.27\,A$
Hence the current drawn by the first lamp is $0.45\,A$ and the current drawn by the second lamp is $0.27\,A$.
Note: Ohm’s law states that the voltage is directly proportional to the current. The first formula of the resistance is derived from the formula $P = VI$ . In this formula, $I = \dfrac{V}{R}$ is substituted as $P = \dfrac{{V \times V}}{R}$. It is rearranged as $R = \dfrac{{{V^2}}}{P}$.
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