Question

Solve the following problems.

A. The resistance of a 1 m long nichrome wire is 6 $\mathrm{\Omega }$. If we reduce the length of the wire to 70 cm. what will its resistance be?

B. When two resistors are connected in series, their effective resistance is 80 $\mathrm{\Omega }$. When they are connected in parallel, their effective resistance is 20 $\mathrm{\Omega }$. What are the values of the two resistances?

C. If a charge of 420 C flows through a conducting wire in 5 minutes what is the value of the current?

Answer 1

A. Resistance of a wire is directly proportional to its length. Let R₁ and R₂ be the resistances of the nichrome wire at 100 cm (1 m) and 70 cm, respectively. Therefore,
$$\begin{gathered} R_1\propto 100.....\left(\mathrm{i}\right) \\ \mathrm{and} \\ {R}_2\propto 70.....\left(\mathrm{ii}\right) \\ \mathrm{Dividing}\left(\mathrm{ii}\right)\mathrm{by}\hspace{0.17em}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get} \\ \frac{R_2}{R_1}=\frac{70}{100} \\ \Rightarrow R_{\mathit{2}}=\frac{70}{100}\times R_1 \end{gathered}$$
Now, R₁ = 6 $\mathrm{\Omega }$
$\Rightarrow R_{\mathit{2}}=\frac{70}{100}\times 6=4.2\mathrm{\Omega }$

B. Let the two resistances be R₁ and R₂. Therefore,
R₁ + R₂ = 80 $\mathrm{\Omega }$
or, R₁ = 80 $-R_2$ .....(i)
and
$$\begin{gathered} \frac{1}{R_1}+\frac{{\displaystyle 1}}{{\displaystyle R_2}}=\frac{{\displaystyle 1}}{20} \\ \Rightarrow R_1{R}_2=20(R_1+R_2) \\ \mathrm{As},R_1=80-R_2 \\ \Rightarrow (80-R_2)R_2=20(80-R_2+R_2) \\ 80R_2^2-R_2=1600 \\ {R}_2^2-80R_2+1600=0 \\ (R_2-40)(R_2-40)=0 \\ \Rightarrow R_2=40\mathrm{\Omega } \\ \mathrm{From}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get} \\ {R}_1=80-40=40\mathrm{\Omega } \end{gathered}$$

c. Current flowing through the conducting wire is
$$\begin{gathered} I=\frac{\mathrm{Charge}\left(Q\right)}{\mathrm{Time}\left(t\right)} \\ \Rightarrow I=\frac{420}{5\times 60}=1.4\mathrm{A} \end{gathered}$$
Hence, the current flowing through the wire is 1.4 A.