Answer:

The new resistor has resistance of 9Ω.

Explanation:

The effective resistance of three resistance is

[tex]$ (1). \: \: \frac{1}{R_e} = \frac{1}{R_1}+ \frac{1}{R_2}+ \frac{1}{R_3} $[/tex].

we know the the circuit with 9V battery has current of 3.6A; therefore, [tex]R_e[/tex] must be

[tex]V = IR_e[/tex]

[tex]R_e = \dfrac{V}{I} = \dfrac{9V}{3.6A}[/tex]

[tex]R_e = 2.5\Omega[/tex],

which means

[tex]$\frac{1}{2.5\Omega } = \frac{1}{R_1}+ \frac{1}{R_2}+ \frac{1}{R_3} $[/tex]

[tex]$\boxed{0.4\Omega^{-1} = \frac{1}{R_1}+ \frac{1}{R_2}+ \frac{1}{R_3} .} $[/tex]

Now, Sally wants to connect another resistor in parallel so that the current becomes 4.6A. The resistance required for that is

[tex]R_{new} = \dfrac{9V}{4.6A}[/tex]

[tex]R_{new} = 1.957 \Omega[/tex].

which means the 4th resistor [tex]R_4[/tex] must satisfy the equation

[tex]$ (2), \: \: \frac{1}{1.957\Omega} = \frac{1}{R_1}+ \frac{1}{R_2}+ \frac{1}{R_3}+\frac{1}{R_4} , $[/tex]

and since

[tex]$ \frac{1}{R_1}+ \frac{1}{R_2}+ \frac{1}{R_3} =0.4\Omega^{-1}, $[/tex]

equation (2) becomes

[tex]$ \frac{1}{1.957\Omega} =0.4\Omega^{-1}+\frac{1}{R_4} , $[/tex]

which we solve for [tex]R_4[/tex]:

[tex]$ 0.511\Omega^{-1} =0.4\Omega^{-1}+\frac{1}{R_4} , $[/tex]

[tex]0.111\Omega^{-1} = \dfrac{1}{R_4}[/tex]

[tex]\boxed{R_4 = 9\Omega}[/tex]

Thus, the resistance of the new resistor is 9 ohms.