Question

35.A wire having length 2m, diameter 1mm and resistivity 1.963 x 10-8
ohm-m is connected in series with battery of emf 3V and internal
resistance 1 ohm. Calculate the resistance of the wire and current in the
circuit.​

Answer 1

$GIVEN:-$

• Length (l) = 2m
• Diameter (D) = 1mm = 1 × 10$^{-3}$
• Radius (r) = $\frac{D}{2}$ = 0.5 × 10$^{-3}$
• Resistivity (f) = 1.963 × 10$^{-8}$ Ωm
• E = 3V
• Internal resistance (r) = 1 Ω

$TO\:FIND:-$

• Resistance of the wire
• Current in the circuit.

$SOLUTION:-$

• $R = \dfrac{fl}{A} = \dfrac{fl}{\pi r^{2}}$

• $A = \pi r^2$

$\implies\:\:R = \dfrac { 1.963 \times 10^{-8} \times 2 }{ 3.14 \times (0.5\times 10^{3})^{2} }$

$\implies\:\:R = \dfrac { 1.963 \times 2\times 10^{-2} }{3.14\times 0.5\times 0.5 }$

$\implies\:\:{\boxed{R = 0.05\Omega}}$

Now, Finding current (I):

$\implies\:\:I = \dfrac{E}{R + r}$

$\implies\:\:I = \dfrac{ 3}{0.05+1}$

$\implies\:\:{\boxed{I = 2.85A}}$