Two wires of the same material have length 6 cm and 10 cm and radii 0.5 mm and 1.5 mm respectively. They are connected in series across a battery of 16 V. The potential difference. across the shorter wire is

A
$5 V$
B
$13.5 V$
C
$27 V$
D
10 V

Solution:

Length $l_{1} = 6 cm$ Length $l_{2} = 10 cm$ Radius $r_{1} = 0.5 mm$ Radius $r_{2} = 1.5 mm$
$V = iR \Rightarrow i \frac{ρℓ}{A} \Rightarrow \frac{V_{1}}{V_{2}} = \frac{ℓ_{1}}{ℓ_{2}} \times {[\frac{r_{2}}{r_{1}}]}^{2} \Rightarrow \frac{27}{5}$

$V_{1} + V_{2} = 16; V_{2} = \frac{5 V_{1}}{27}$

$\Rightarrow V_{1} + \frac{5 V_{1}}{27} = 16 \Rightarrow V_{1} = \frac{16 \times 27}{32} = 13.5 V$

Two wires of the same material have length 6 cm and 10 cm and radii $0 .5 mm$ and $1 .5 mm$ respectively. They are connected in series across a battery of $16 V$ . The potential difference. across the shorter wire is

Solution:

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