Question

An electric lamp of $100 \Omega$, a toaster of resistance $50 \Omega$ and a water filter of resistance $500 \Omega$ are connected in parallel to a 220V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances and what is the current through it?

Answer 1

Hint: To solve this question, use Ohm’s law. Firstly, find the current flowing through all the three appliances i.e. electric lamp, toaster and a water filter individually. Sum of the current of all these three appliances will give the total current flowing through the electric iron. Thus, add all the currents to get the total current. Using this obtained total current and given source voltage, calculate the resistance of an electric current.

Complete answer:
Given: Voltage (V)= 220V
             Resistance through electric lamp= $100 \Omega$
             Resistance through toaster= $50 \Omega$
             Resistance through water filter= $500 \Omega$
According to Ohm’s law.
$V=IR$
Where, V is the potential difference
             I is the current flowing
             R is the resistance
Rearranging above equation we get,
$I= \dfrac {V}{R}$
Thus, current I flowing through electric lamp can be calculated by substituting values in above equation, we get,
${I}_{1}= \dfrac {220}{100}$
$\Rightarrow{I}_{1}= \dfrac {22}{10}$
Similarly, current through the toaster is,
${I}_{2}= \dfrac {220}{50}$
$\Rightarrow{I}_{2}= \dfrac {22}{5}$
Current through the water filter is given by,
${I}_{3}= \dfrac {220}{500}$
$\Rightarrow{I}_{3}= \dfrac {22}{50}$
Thus, the total current is,
$Total \quad current= { I}_{1}+{ I}_{2}+{ I}_{3}$
Substituting the values we get,
$Total \quad current= \dfrac {22}{10}+ \dfrac {22}{5}+\dfrac {22}{50}$
$\Rightarrow Total \quad current= \dfrac {22}{5} (\dfrac {1}{2}+1+\dfrac {1}{10})$
$\Rightarrow Total \quad current= \dfrac {22}{5} \times \dfrac {8}{5}$
$\Rightarrow Total \quad current= \dfrac {176}{25}$
$\Rightarrow Total \quad current= 7.04A$
Resistance of the electric iron can be calculated by,
$R= \dfrac {V}{I}$
Substituting values in above equation we get,
$R= \dfrac {220}{7.04}$
$\therefore R= 31.25 \Omega$
Therefore, resistance of the electric iron is $31.25\Omega$ and current flowing through it is $7A$.

Note:
Resistance depends on the type of material as well as its dimensions i.e. area of cross-section. Materials offering very high resistance are called Insulators. While, the materials offering very low resistance are called Conductors. Greater the value of resistance, smaller is the flow of current through the material. The relation between applied voltage and current depends on the sign of the voltage. If I is the current for a certain value of voltage V, then reversing the direction of V does not produce the current of same magnitude as the I in the opposite direction.