Exercise 266 Part A For the circuit shown in the figure Figu

Exercise 26.6 Part A For the circuit shown in the figure (Figure 1) both meters are idealized, the battery has no appreciable internal resistance, and the ammeter reads 1.15 A What does the voltmeter read? Submit My Answers Give Up Part B What is the emf E of the battery? Figure 1 of 1 Submit My Answers Give Up 25.0 15.0 15.0 Provide Feedback Continue 45.0 10.0 35.0 ?

Solution

26.6:

The potential difference on R2 : V₂= I₂R₂ =1.15*25 = 28.75 V

So the current in R3 is,

      I3 = V2/R3 = 28.75/15= 1.9167 A

the current in R4 IS,

       I4 = V2/(R4+10)

           =28.75/[15+10]

           =1.15 A

Hence, the total current in the circuit is,

          I=I2+I3+I4 =1.15+1.9167+1.15 = 4.2167 A

therefore, the voltmeter read:

      V1= I.R1 = 4.2167 * 45 = 189.7515 V = 189.75 V

the emf of the battery is,

     =V2+V1+I.(35)= 28.75 + 189.75 + 4.2167*35 = 366.086 V