Electricity Test - 63

• Two resistors when connected in parallel always have equivalent resistance less than or equal to the resistance of the smaller resistor.  $$\\ \dfrac 1R_{eq} = \dfrac 1R_1 + \dfrac 1R_2$$
• Since the smaller resistance is $$6 \Omega$$ ,the resistance of $$R_1$$ has to be greater than or equal to $$6 \Omega$$.
  

Given that,

The dimension of the block is 1cm, 3cm and 3cm.

The relation of the resistance of a conductor is

$$R=\rho \dfrac{l}{A}....(I)$$

Where,

$$\rho $$ = resistivity of the material of the conductor

$$A$$= area of the cross section of the conductor

$$l$$= length of the conductor

Now, the resistance is maximum

$$ {{R}_{\max }}=\rho \times \dfrac{3}{1\times 2} $$

$$ {{R}_{\max }}=\dfrac{3\rho }{2}...(II) $$

Now, the resistance is minimum

$$ {{R}_{\min }}=\rho \dfrac{1}{2\times 3} $$

$$ {{R}_{\min }}=\dfrac{\rho }{6}...(III) $$

Now, from equation (II) and (III)

The ratio of maximum resistance and minimum resistance is

$$ \dfrac{{{R}_{\max }}}{{{R}_{\min }}}=\dfrac{\frac{3\rho }{2}}{\dfrac{\rho }{6}} $$

$$ \dfrac{{{R}_{\max }}}{{{R}_{\min }}}=\dfrac{3}{2}\times \dfrac{6}{1} $$

$$ \dfrac{{{R}_{\max }}}{{{R}_{\min }}}=\dfrac{9}{1} $$

$$ {{R}_{\max }}:{{R}_{\min }}=9:1 $$

Hence, the ratio is $$9:1$$

Resistance in wire, $$R=\rho \dfrac{L}{A}=\dfrac{\rho L}{\pi {{r}^{2}}}$$

  $$ {{R}_{1}}=\dfrac{\rho {{L}_{1}}}{\pi r_{1}^{2}}=\dfrac{\rho {{L}_{1}}}{\pi {{(2r)}^{2}}}=\dfrac{1}{4}\dfrac{\rho {{L}_{1}}}{\pi {{r}^{2}}}=34\,\Omega  $$

 $$ \Rightarrow \dfrac{\rho {{L}_{1}}}{\pi {{r}^{2}}}=34\times 4=136\ \Omega \ ......\ (1) $$

$$L_2=L_1/2$$      $$r_2=r$$

 $$ {{R}_{2}}=\rho \dfrac{{{L}_{2}}}{{{A}_{2}}}=\dfrac{\rho \dfrac{{{L}_{1}}}{2}}{\pi r_{2}^{2}}=\dfrac{1}{2}\dfrac{\rho {{L}_{1}}}{\pi {{r}^{2}}}=\dfrac{1}{2}\times 136=\ 68\,\Omega  $$

Resistance of B is $$68\,\Omega .$$