Question

an electric lamp of 60 ohm &toaster of 30 ohms are connected in parallel to a 20 v source of electricity a ) what will be the resistance of an electric iron which when connected to the same electric source permits the flow of the same electric current as by the arrangement of the 2 appliances stated above
b) what is the current passing through the electric iron
c) calculate the power of electric iron

Answer 1

$$\begin{gathered} \mathrm{Given} \\ \mathrm{Resistance}\mathrm{of}\mathrm{electric}\mathrm{lamp},R_1=60\mathrm{\Omega } \\ \mathrm{Resistance}\mathrm{of}\mathrm{toaster},R_2=30\mathrm{\Omega } \\ \mathrm{As}\mathrm{they}\mathrm{are}\mathrm{connected}\mathrm{in}\mathrm{parallel},\mathrm{net}\mathrm{resistance}\mathrm{becomes} \\ \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2} \\ \Rightarrow \frac{1}{R}=\frac{1}{60}+\frac{1}{30} \\ \Rightarrow \frac{1}{R}=\frac{90}{1800} \\ \Rightarrow R=20\mathrm{\Omega } \\ \mathrm{Now},\mathrm{current}\mathrm{flowing}\mathrm{through}\mathrm{the}\mathrm{circuit}\left(\mathrm{I}\right)\mathrm{is}\mathrm{given}\mathrm{by} \\ I=\frac{V}{R}=\frac{20}{20} \\ \Rightarrow I=1\mathrm{A} \\ \left(\mathrm{a}\right) \\ \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{question},\mathrm{same}\mathrm{current}\mathrm{should}\mathrm{flow}\mathrm{through}\mathrm{the}\mathrm{iron}.\mathrm{Therefore}\mathrm{its} \\ \mathrm{Reistance},R=\frac{V}{I} \\ \Rightarrow R=\frac{20}{1}=20\mathrm{\Omega } \\ \left(\mathrm{b}\right)\mathrm{Current}\mathrm{through}\mathrm{the}\mathrm{iron}\mathrm{is}1\mathrm{A}. \\ \left(\mathrm{c}\right)\mathrm{Power}\mathrm{of}\mathrm{the}\mathrm{iron},\mathit{}P=I^{\mathit{2}}R \\ \Rightarrow P={\left(1\right)}^2\times 20=20\mathrm{watt} \end{gathered}$$