Question

20) Equivalent resistance of
series combination of two
conducting wires is 14 ohm and of
parallel combination of them is
3.43 ohm so resistance of wire
having greater magnitude​

Answer 1

Explanation:

20) Equivalent resistance of

series combination of two

conducting wires is 14 ohm and of

parallel combination of them is

3.43 ohm so resistance of wire

having

Answer 2

The resistance of wire having greater magnitude, R2(R2>R1) is $\frac{48.02}{R1} ohm$.

Given,

Equivalent resistance of series combination of two conducting wires=14 ohm

Equivalent resistance of parallel combination of two conducting wires=3.43 ohm.

To find,

the resistance of wire having greater magnitude.

Solution:

• When resistors form a parallel combination, the equivalent resistance comes to be as per the following expression:
• $\frac{1}{Rnet} =\frac{1}{R1} +\frac{1}{R2}$.
• When resistors form a series combination, the equivalent resistance comes to be as per the following expression:
• $Rnet=R1+R2$.
• Voltage supplied by the battery is same in each resistor when the resistors are in a parallel combination. Its the current that gets divided.
• Current supplied by the battery is same in each resistor when the resistors are in a parallel combination. Its the voltaget that gets divided.

Let the two resistors be R1 and R2. And let us assume that R2 is greater than R1.

The expression for equivalent resistance of parallel combination will be:

$\frac{1}{Rnet} =\frac{1}{R1} +\frac{1}{R2}\\\frac{1}{14} =\frac{1}{R1} +\frac{1}{R2}\\\frac{1}{14} =\frac{R2+R1}{R1R2} \\R1R2=14(R1+R2).$

The expression for equivalent resistance of series combination will be:

$Rnet=R1+R2\\3.43=R1+R2.$

As the value of (R1+R2) comes as 3.43 ohm from the expression above, the value of R1.R2 will be:

$R1R2=14(R1+R2)\\R1R2=14(3.43)\\R1R2=48.02\\R2=\frac{48.02}{R1} ohm.$

Hence, the value of R2(R2>R1) is $\frac{48.02}{R1} ohm$.

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