In the circuit shown in Fig. E26.31 the batteries have negligible internal resistance and the meters are both idealized. With the switch Sopen, the voltmeter reads (a) Find the emf E of the battery. (b) What will the ammeter read when the

switch is closed?

Short Answer

Expert verified

The EMF of the Battery is 36.4V

the ammmeter raeding when switch is closed is 0.500A

Step by step solution

01

About EMF of the battery

Electromotive forceis defined as the electric potential produced by either electrochemical cell or by changing the magnetic field

02

Determine the EMF of the battery 

Before attacking part a, We can make some simplifications to the circuit to make our job a little easier. Recognize that thebatteries have no internal resistance, the meters are idealized, and with the switch open, the voltmeter reads 15.0 V.At 50.Ohms

50Ω=15Vl=1550=0.300A

The 50ohmsand 30.ohmsresistors are in series, so We can combine themThere will be the same current running through thescWe also know that the voltage at 80.ohms = the voltage at 75.ohms due to the parallel nature of the resistors.

V=IR=0.300A×80Ω=24V

The 75.ohms resistor has a different current than We've calculated, but we can use the voltage We just determined-

03

:Determine the reading of ammeter

l=2475=0.320Alatl^=0.300A+0.320A

For the 75ohms and 80ohms resistor

Req=175+180=38.7Ω

The battery has no internal resistance, but it is in series with the resistors in paralleL

38.7+20=58.7V=IR=0.620A×58.7=36.4V

Therefore The EMF of the Battery is 36.4V

04

:Determine when the switch is closed

b.)withtheswitchclosed,theloopiscompletedforthe25.0Vbatteryandthe50.ohmsresistor.
V = IR

25V=l50.00l=2550=0.500A

therefore the ammmeter raeding when switch is closed is 0.500A
Result
a) 36.4 V, b.) 0.500 A

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