Question

1. Write the formula for resistance of a wire of length L and cross-section A.
2. A piece of wire is redrawn by pulling it until its length is doubled. Compare the new
resistance with the original value.
3. How many 176$\Omega$ resistors (in parallel) are required to carry 5 A in 220 V line? [5 MARKS]

Answer 1

Equation: 1 Mark
Each problem: 2 Marks

1. R = ρ$\frac{L}{A}$
2. Volume of the material of wire remains same. So, when length is doubled, its area of cross - section will get halved. So, if l and a are the original length and area of cross-section of wire,

Original value of the resistance, $R=\rho \times \frac{l}{a}$

and,

New value of the resistance,

$R^{{}^{\prime }}=\rho \times \frac{2l}{\frac{a}{2}}=\rho \frac{1}{a}\times 4=4R$

3. Here, I = 5A, V = 220 V.

Resistance required in the circuit, R = $\frac{V}{I}$ = $\frac{220V}{5A}$ = 44 $\Omega$, Resistance of each resistor, r = 176 $\Omega$

If n resistors, each of resistance r, are connected in parallel to get the required resistance R, then R = $\frac{r}{n}$ or n = $\frac{r}{R}$ = $\frac{176}{44}$ = 4