The resistance of a wire that must be placed parallel with a 12 Q resistance to obtain a combined resistance of 4 Q is

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Q: 100 (NDA-I/2009)
The resistance of a wire that must be placed parallel with a 12 Q resistance to obtain a combined resistance of 4 Q is

question_subject: 

Science

question_exam: 

NDA-I

stats: 

0,5,4,2,1,5,1

keywords: 

{'combined resistance': [0, 0, 2, 0], 'resistance': [0, 0, 1, 2], 'wire': [0, 0, 7, 16]}

To obtain a combined resistance of 4 Ω by placing a wire in parallel with a 12 Ω resistance, we need to find the resistance of the wire that needs to be added.

When resistances are connected in parallel, the formula for calculating the combined resistance is:

1/Rcombined = 1/R1 + 1/R2 + ...

Let`s substitute the given values into the formula and solve for the resistance of the wire:

1/Rcombined = 1/12 + 1/Rwire

We need to rearrange the equation to solve for Rwire:

1/Rwire = 1/Rcombined - 1/12

Combining the fractions on the right side:

1/Rwire = (12 - Rcombined) / (12 * Rcombined)

To simplify the equation, we can multiply both sides by Rwire * 12:

12 = (12 - Rcombined) * Rwire

Expanding the equation:

12 = 12Rwire - Rcombined * Rwire

Since we know that the combined resistance is 4 Ω, we can substitute it into the equation:

12 = 12Rwire - 4Rwire

Simplifying the equation:

12 = 8Rwire

Now we can solve for Rwire