A circuit is constructed with five resistors and a battery a

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 = 58 ?, R2 = 105 ?, R3 = 122 ?, and R4 = 132 ?. The value for RX is unknown, but it is known that I4, the current that flows through resistor R4, is zero.

Circuit 1 with Resistors and a Battery 1 2 34567 circuit is constructed with five resistors and a battery as shown. The 105 , R3 122 , and R-132 . The value for Rx is unknown, but it battery voltage is 12 V. The values for the resistors are: R-58 , R2 Ri is known that l4, the current that flows through resistor R4, is zero 1)What is , the magnitude of the current that flows through the resistor R 2) what is vz, the magnitude of the voltage across the resistor R2? 3) what is ,the magnitude of the current that flows through the resistor R? Sutarmit 4) what is Rx,the value of the unknown resistor R init )what is v,, the magnitude of the voltage across the resistor R 5) if the value of the resistor Rz were doubled, how would the value of the resistor R have to change in order to keep the current through R4 equal to zero? O R would need to be increased O R3 would need to be decreased O Ry would not need to be changed O There is no change that could be made to R to keep the current through R4 equal to zero.

Solution

a)

Given Current through R₄ is zero ,so using the principles of wheat stones bridge under balanced conditons

R₃/R₁ = R_x/R₂

=>R_x = (R₃/R₁)*R₂ =(122/58)*105

R_x = 220.86 ohms

Voltage across R1

V₁=V*(R₁/R₁+R₃) =12*(58/58+122) =3.87 V

Current through R1

I₁=V₁/R₁  =3.87/58 =0.0667 A

b)

Voltage acorss V₂

V₂=V*(R₂/R₂+R_x) =12*(105/105+220.86)

V₂=3.87 Volts

c)

Current throught R2 is

I₂=V₂/R₂=3.87/105 =0.0368 A

d)

R_x=220.86 ohms

e)

Voltage across R₁

V₁=3.87 V

f)

Given R₂ = 2*105 =210 ohms

From bridge balanced condition

R₃/R₁ =R_x/R₂

R₃ =58*(220.86/210) =61 ohms

R3 would need to be decreased .