Answer:
[tex]214.9 \Omega[/tex]
Explanation:
The three resistors are connected in parallel: this means that the potential difference across each resistor is the same as the voltage of the battery. This can be calculated using the information about the [tex]38.7 \Omega[/tex] resistor: in fact, since we know its resistance and the current flowing through it (0.155 A), we can find the potential difference across this resistor, which is equal to the voltage of the battery:
[tex]V=IR=(0.155 A)(38.7 \Omega)=6.0 V[/tex]
We also know the total current in the circuit, 0.250 A. This means that we can find the total resistance of the circuit, using Ohm's law:
[tex]R_{eq}=\frac{V}{I}=\frac{6.0 V}{0.250 A}=24 \Omega[/tex]
So now we now the total resistance and the resistance of two of the 3 resistors; therefore, we can find the resistance of the 3rd resistor:
[tex]\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\\\frac{1}{R_3}=\frac{1}{R_{eq}}-\frac{1}{R_1}-\frac{1}{R_2}=\frac{1}{24 \Omega}-\frac{1}{38.7\Omega}-\frac{1}{89.5\Omega}=0.00465 \Omega^{-1}\\R_3=\frac{1}{0.00465 \Omega^{-1}}=214.9 \Omega[/tex]
The resistances arranged in parallel reduce the net resistance of circuit. The magnitude of the resistance of the third resistor is of [tex]214.9 \;\rm \Omega[/tex].
The hindrance offered to the flow of electrons through any cross-section area is known as the electrical resistance.
Given data -
The resistances of two resistors are, [tex]38.7 \;\rm \Omega[/tex] and [tex]89.5 \;\rm \Omega[/tex].
The current through [tex]38.7 \;\rm \Omega[/tex] resistor is, I = 0.155 A.
The total circuit current is, I' = 0.250 A.
Let us first calculate the potential difference across [tex]38.7 \;\rm \Omega[/tex] resistor. Then, by Ohm's law,
[tex]V = I \times R_{1}\\\\V =0.155 \times 38.7\\\\V = 6.0 \;\rm V[/tex]
Now, the total resistance of circuit is,
[tex]V = I' \times R'\\\\R' = \dfrac{V}{I'}\\\\R' = \dfrac{6}{0.250}\\\\R' = 24 \;\rm \Omega[/tex]
With parallel combination, the third resistance is obtained using the expression,
[tex]\dfrac{1}{R'}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}\\\\\dfrac{1}{R_{3}}=\dfrac{1}{R'}-\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}}[/tex]
Solving as,
[tex]\dfrac{1}{R_{3}}=\dfrac{1}{24}-\dfrac{1}{38.7}-\dfrac{1}{89.5}\\\\R_{3} = 214.9 \;\rm \Omega[/tex]
Thus, we can conclude that the magnitude of the resistance of the third resistor is of [tex]214.9 \;\rm \Omega[/tex].
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