Question

Three resistors are wired in parallel with a battery. Two of the resistors have resistances of 38.7 Q/ and 89.5 Q. The current in the 38.7 Q resistor is 0.155 A and the total circuit current Is 0.250 A. What is the resistance of the third resistor?

Answer 1

Answer:

$214.9 \Omega$

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 $38.7 \Omega$ 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:

$V=IR=(0.155 A)(38.7 \Omega)=6.0 V$

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:

$R_{eq}=\frac{V}{I}=\frac{6.0 V}{0.250 A}=24 \Omega$

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:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\\\frac{1}{R_3}=\frac{1}{R_{eq}}-\frac{1}{R_1}-\frac{1}{R_2}=\frac{1}{24 \Omega}-\frac{1}{38.7\Omega}-\frac{1}{89.5\Omega}=0.00465 \Omega^{-1}\\R_3=\frac{1}{0.00465 \Omega^{-1}}=214.9 \Omega$