Answer:
1160 ohm
Explanation:
We are given that
R'=580 ohm
Current=3 I
We have to find the resistance of the circuit.
Let R be the resistance of circuit.
In parallel
[tex]\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}[/tex]
Using the formula
[tex]\frac{1}{R_{eq}}=\frac{1}{580}+\frac{1}{R}=\frac{580+R}{580R}[/tex]
[tex]R_{eq}=\frac{580R}{580+R}[/tex]
In parallel combination,Potential difference across each resistance remains same.
[tex]V=IR[/tex]
Using the formula
[tex]IR=3IR_{eq}[/tex]
[tex]IR=3I\times \frac{580R}{580+R}[/tex]
[tex]580+R=3\times 580=1740[/tex]
[tex]R=1740-580=1160\Omega[/tex]