Question

two wires when connected in series have an equivalent resistance of 18 ohm and when connected in parallel an equivalent resistance of 4 ohm. find their remittances?
A-10 ohm, 33ohm
B-12 ohm, 6ohm
C-26ohm,2ohm​

Answer 1

Answer:

(B) R1=12Ω & R2=6Ω

Explanation:

In series combination,

R1+R2=18Ω——————(1)

In parallel combination,

$\frac{1}{R1} + \frac{1}{R2 } = \frac{1}{4} \\ \frac{R1R2 }{R1+ R2 } = 4 \\ R1R2 = 4 \times 18 = 72 \\ R1 = \frac{72}{R2 }$

$\frac{72}{R2 } + R2 = 18 \\ 72 + {R2 }^{2} = 18R2 \\ {R2 }^{2} - 18R2 + 72 = 0 \\ {R2 r2}^{2} - 12R2 - 6R2 + 72 = 0 \\ R2 (R2 - 12) - 6(R2- 12) = 0 \\ R2 = 12 \: R2 = 6$

R2=6Ω, R1=12Ω

Answer 2

The value of resistances are 4 ohms and 12 ohms.

Explanation:

Given that, two wires when connected in series have an equivalent resistance of 18 ohm and when connected in parallel an equivalent resistance of 4 ohm.

Let $R_1\ and\ R_2$ are two resistances. In series combination, the equivalent resistance is given by :

$R=R_1+R_2$

$18=R_1+R_2$............(1)

In parallel combination, the equivalent resistance is given by :

$\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}$

$\dfrac{1}{4}=\dfrac{1}{R_1}+\dfrac{1}{R_2}$.........(2)

On solving equation (1) and (2) we get :

$R_1=6\ \Omega$

$R_2=12\ \Omega$

So, the value of resistances are 4 ohms and 12 ohms. Hence, this is the required solution.

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