A circuit is constructed with five resistors and a battery as shown. the battery

Question

A circuit is constructed with five resistors and a battery as shown. the battery voltage is V = 12 V. the values for the resistors are: R_1 = 66 Ohm, R_2 = 141 Ohm, R_3 = 114 Ohm, and R_4 = 76 Ohm. the value for R_x is unknown, but it is known that l_4, the current that flows through resistor R_4 is zero. What is 1_1, the magnitude of the current that flows through the resistor R_1? What is V_2 the magnitude of the voltage across the resistor R_2? What is l_2, the magnitude of the current that flows through the resistor R_2? What is R_x, the value of the unknown resistor R_x? What is V_1, the magnitude of the voltage across the resistor R_1? If the value of the resistor R_z was doubled, how would the value of the resistor R_3 have to change in order to keep the current through R_4 equal to zero? R_3 would need to be increased

Explanation / Answer

1) I1 = V/(R1+R3)

12/(66+141) = 0.0579 A

2)the voltage is equal to V1
  
V1 = R1 * I1

0.0579 * 66 = 3.8214

V2 = 3.8214 V

3) I2 = V2/R2

3.8214/141 = 0.0271

I2 = 0.0271 A

4) 12 = I2(R2+Rx)

Rx = (12-0.0271*141)/0.0271

Rx = 301.8044

5)

V1 = R1*I1

= 0.0579*66 = 3.8214

V1 = 3.8214 V

6) If R2 is doubled keeping R1, R4 and Rx fixed, from the balancing condition =
  
= R3'/R1 = Rx/R2'

= R3' = R1 * (Rx / 2*R2) = (1/2) * (R1 * Rx / R2) = (1/2) * R3

R3 should be half its initial value (decreased)

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