Two resistors are joined in parallel having equivalent resistance 6/5 ...
Let first resistance be r1 and second resistance be r2
then, 1/r1 + 1/r2 = 1/(6/5) = 5/6
=》1/r1 + 1/r2 = 5/6 ....(1)
when when of the resistance breaks then the effective resistance becomes 2 ohm.Let r1 is broken then the effective resistance will be equal to the resistance of r2.
so, r2 = 2 ohm
putting r2 = 2 in (1),
we get
1/r1 + 1/r2 = 5/6
=》1/r1 + 1/2 = 5/6
=》1/r1 = 5/6 - 1/2 = 2/6 = 1/3
=》1/r1 = 1/3
=》r1 = 3
r1 is the broken resistance and its resistance is 3 ohm which can also be said to be the resistance of broken resistor.
So, resistance of broken resistor = 3 ohm
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Two resistors are joined in parallel having equivalent resistance 6/5 ...
Understanding the Problem
When two resistors are connected in parallel, the formula for equivalent resistance (R_eq) is given by:
1/R_eq = 1/R1 + 1/R2
Here, we know the equivalent resistance R_eq = 6/5 ohm.
Calculating Individual Resistances
Let R1 be the resistance of the first wire, and R2 be the resistance of the second wire. Using the given equivalent resistance, we can write:
1/(6/5) = 1/R1 + 1/R2
This simplifies to:
5/6 = 1/R1 + 1/R2
By multiplying through by R1 * R2, we get:
5R1R2 = 6(R1 + R2)
Now, we need to find out what happens when one of the resistors breaks.
Resistance After Breakage
If one resistor (let's say R2) breaks, the effective resistance becomes 2 ohms. The remaining resistor R1 will solely define the equivalent resistance now:
R_eq (after break) = R1 = 2 ohm
Finding the Value of R2
Now that we know R1 = 2 ohm, we can substitute back into our earlier equation:
5R1R2 = 6(R1 + R2)
Substituting R1 = 2:
5(2)R2 = 6(2 + R2)
This results in:
10R2 = 12 + 6R2
Rearranging gives:
4R2 = 12
Thus, R2 = 3 ohm.
Conclusion
The resistance of the wire that broke is:
- R2 = 3 ohm
This analysis shows how to derive the resistance values based on the given conditions.
Two resistors are joined in parallel having equivalent resistance 6/5 ...
3ohm
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