Question

Sally Sizzle builds a circuit by connecting three resistors in oarallel with a 9v battery. The current in this curvuit is 3.6 amps. Sally wants to add another resistor in parallel tonincrrase the current to 4.6 amps. What should the resistance of this new resistor be?

Answer 1

Answer:

The new resistor has resistance of 9Ω.

Explanation:

The effective resistance of three resistance is

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

we know the the circuit with 9V battery has current of 3.6A; therefore, $R_e$ must be

$V = IR_e$

$R_e = \dfrac{V}{I} = \dfrac{9V}{3.6A}$

$R_e = 2.5\Omega$,

which means

$\frac{1}{2.5\Omega } = \frac{1}{R_1}+ \frac{1}{R_2}+ \frac{1}{R_3}$

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

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

$R_{new} = \dfrac{9V}{4.6A}$

$R_{new} = 1.957 \Omega$.

which means the 4th resistor $R_4$ must satisfy the equation

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

and since

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

equation (2) becomes

$\frac{1}{1.957\Omega} =0.4\Omega^{-1}+\frac{1}{R_4} ,$

which we solve for $R_4$:

$0.511\Omega^{-1} =0.4\Omega^{-1}+\frac{1}{R_4} ,$

$0.111\Omega^{-1} = \dfrac{1}{R_4}$

$\boxed{R_4 = 9\Omega}$

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