91Ó°ÊÓ

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

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

$$\begin{gathered} 50\mathrm{Î\copyright }=15\mathrm{V} \\ \mathrm{l}=\frac{15}{50}=0.300\mathrm{A} \end{gathered}$$

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.

$$\begin{gathered} \mathrm{V}=\mathrm{IR} \\ =0.300\mathrm{A}×80\mathrm{Î\copyright } \\ =24\mathrm{V} \end{gathered}$$

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

$$\begin{gathered} \mathrm{l}=\frac{24}{75}=0.320\mathrm{A} \\ \mathrm{lat}\hat{\mathrm{l}}=0.300\mathrm{A}+0.320\mathrm{A} \end{gathered}$$

For the 75ohms and 80ohms resistor

$$\begin{gathered} {\mathrm{R}}_{\mathrm{eq}}=\frac{1}{75}+\frac{1}{80} \\ =38.7\mathrm{Î\copyright } \end{gathered}$$

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

$$\begin{gathered} 38.7+20=58.7 \\ \mathrm{V}=\mathrm{IR} \\ =0.620\mathrm{A}×58.7 \\ =36.4\mathrm{V} \end{gathered}$$

Therefore The EMF of the Battery is 36.4V

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

$$\begin{gathered} 25\mathrm{V}=\mathrm{l}50.00 \\ \mathrm{l}=\frac{25}{50}=0.500\mathrm{A} \end{gathered}$$

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