Question

Two coils are connected in parallel and a voltage of 200 V is applied to the terminals. The total current taken is 25 A and the power dissipated in one of the coil is 1500 W. Determine the resistance of each coil.

Answer 1

Given:-

• Voltage(V) = 200V
• Total current (It) = 25A
• Power in first coil(P1) = 1500W

We have to find,

• Resistance of second coil(R2) = ?

Solution :-

Since it’s a parallel connection, voltage is same for both coils.

R = V/It = 200/25 = 8 Ohm.

Now,

$\sf \longmapsto \: P_{1} \: = \: \dfrac{V {}^{2} }{ R_{1} } \\ \\ \sf \longmapsto R_{1} \: = \: \dfrac{V {}^{2} }{ P_{1} } \\ \\ \sf \longmapsto \: R_{1} = \dfrac{(200) {}^{2} }{1500} \\ \\ \sf \longmapsto \: R_{1} = \dfrac{200 \times 2 \cancel{00}}{15 \cancel{00} } \\ \\ \sf \longmapsto \: R_{1} = \dfrac{80}{3}$

Then, as we know that,

$\sf \longmapsto \: \dfrac{1}{ R_{1} } + \dfrac{1}{ R_{2} } = R \\ \\ \sf \longmapsto \: \: \dfrac{3}{80 } + \dfrac{1}{ R_{2} } = \dfrac{1}{8} \\ \\ \sf \longmapsto \dfrac{1}{ R_{2} } = \dfrac{1}{ 8} - \dfrac{3}{80} \\ \\ \sf \longmapsto \dfrac{1}{ R_{2} } = \dfrac{7}{80} \\ \\ \sf \longmapsto \boxed{ \boxed{ \bf{R_{2} = \frac{80}{7} \: \Omega }}}$

Therefore, the resistance(R2) of other coil is 80/7 ohm.

Answer 2

$\; \; \; \; \; \; \; \; \; \; \; \; \; \;{\Large{\bold{\bf{Required \; answer}}}}$

${\large{\bold{\rm{\underline{Given \; that}}}}}$

✠ Two coils are connected in parallel and a voltage of 200 V is applied to the terminals. Means voltage V = 200V

✠ The total current I taken is 25 A

✠ The power P dissipated in one of the coil is 1500 W.

${\large{\bold{\rm{\underline{To \; determine}}}}}$

✠ The resistance of each coil.

${\large{\bold{\rm{\underline{Solution}}}}}$

✠ The resistance R of each coil = 80/7 Ω

${\large{\bold{\rm{\underline{Full \; Solution}}}}}$

~ As we already see that the question says that the two coils are connected in parallel and a voltage of 200 V is applied to the terminals.

~ Henceforth,

⇢ R = V/I

⇢ R = 200/25

⇢ R = 40/5

⇢ R = 8 Ω

~ Now,

⇢ ${\sf{P_{1} \: = \dfrac{V^{2}}{R_{1}}}}$

• (÷ = ×) ; (× = ÷)

⇢ ${\sf{R_{1} \: = \dfrac{V^{2}}{P_{1}}}}$

⇢ ${\sf{R_{1} \: = \dfrac{200^{2}}{1500}}}$

⇢ ${\sf{R_{1} \: = \dfrac{40000}{1500}}}$

⇢ ${\sf{R_{1} \: = \dfrac{400}{15}}}$

⇢ ${\sf{R_{1} \: = \dfrac{80}{3}}}$

~ As we know that in parallel resistance,

⇢ ${\sf{\dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} = R}}$

⇢ ${\sf{\dfrac{3}{80} + \dfrac{1}{1_{R}} = \dfrac{1}{8}}}$

• (+ = -) ; (- = +)

⇢ ${\sf{\dfrac{1}{R_{2}} = \dfrac{1}{8} - \dfrac{3}{80}}}$

• Let us take the LCM

⇢ ${\sf{\dfrac{1}{R_{2}} = \dfrac{7}{80}}}$

⇢ ${\sf{R^{2} = \dfrac{80}{7}}}$ Ω

Henceforth, 80/7 Ω is the resistance of each coil.