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: R1 = 41 ?, R2 = 152 ?, R3 = 176 ?, and R4 = 124 ?. The value for RX is unknown, but it is known that I4, the current that flows through resistor R4, is zero.

1) What is I1, the magnitude of the current that flows through the resistor R1?

2) What is V2, the magnitude of the voltage across the resistor R2?

3) What is I2, the magnitude of the current that flows through the resistor R2?

4) What is RX, the value of the unknown resistor RX?

5) What is V1, the magnitude of the voltage across the resistor R1?

6)If the value of the resistor R2 were doubled, how would the value of the resistor R3 have to change in order to keep the current through R4 equal to zero?

a) R3 would need to be increased

b) R3 would need to be decreased

c) R3 would not need to be changed

d) There is no change that could be made to R3 to keep the current through R4 equal to zero.

Explanation / Answer

1) As I4 is 0,system behaves as (R1+R3) parallel to (R2+Rx) current through R1 and R3=I1= V/(R1+R3) =12/(217) =0.055 AMPERES 2) As I4 is 0,system behaves as (R1+R3) parallel to (R2+Rx) current through R2 and Rx=I2= V/(R2+Rx) =12/(152+652) .............check value of Rx from bit (4) = 0.015 Voltage across R2=0.015X152 = 2.28 VOLTS 3)As I4 is 0,system behaves as (R1+R3) parallel to (R2+Rx) current through R2 and Rx=I2= V/(R2+Rx) =12/(152+652) .............check value of Rx from bit (4) = 0.015 AMPERES 4) for I4, the current that flows through resistor R4 to be zero,the system of resistors must follow WHEATSTONE BRIDGE (R1/R3)=(R2/Rx) Rx=652 ohms 5) voltage across R1=0.055X 41 =2.255 VOLTS 6) (R1/R3)=(R2/Rx) if R2 is doubled, R3 must be halved Therefore R3 will decrease

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