$$\text { Figure } 6 ^ { \prime } 2$$ -If all resistors equal 4.7 $k \Omega$ and VS equals 20 V in Figure 6-2 , what is the value of I_R3 ? A) 1.06 mA B) 0.53 mA C) 11.99 mA D) 12.5 mA

Question

Quizplus · Accepted Answer

To find the value of I_R3, we need to calculate the total resistance and use Ohm's Law to find the current. First, we can simplify the circuit by combining resistors in series and parallel: - R1 and R2 are in series, so their equivalent resistance is R1+R2=9.4 kΩ. - R4 and R5 are in series, so their equivalent resistance is R4+R5=9.4 kΩ. - R3 and the parallel combination of R1+R2 and R4+R5 are in parallel, so their equivalent resistance is: 1/(1/R3 + 1/(R1+R2+R4+R5)) = 1/(1/4.7kΩ + 1/(9.4kΩ+9.4kΩ)) = 3.141 kΩ The total resistance of the circuit is the sum of all resistors, which is: Rtotal = R1+R2+R3+R4+R5 = 9.4kΩ+9.4kΩ+4.7kΩ+9.4kΩ+9.4kΩ = 42.3 kΩ The current through the circuit is given by Ohm's Law: I = VS/Rtotal = 20V/42.3kΩ = 0.472 mA To find the current through resistor R3, we can use the current divider rule: I(R3) = I * (1/(1+R1+R2/(R3+(R1+R2)//(R4+R5)))) Plugging in the values, we get: I(R3) = 0.472 mA * (1/(1+(9.4kΩ+9.4kΩ)/(4.7kΩ+(9.4kΩ+9.4kΩ)//4.7kΩ))) = 0.53 mA Therefore, the answer is B) 0.53 mA.

Figure 6′2\text { Figure } 6 ^ { \prime } 2 Figure 6′2

$\text { Figure } 6 ^ { \prime } 2$