# Problem: Learning Goal: To understand how to compute power dissipation in a resistive circuit.(Figure 1) The circuit in the diagram consists of a battery with EMF E, a resistor with resistance R, an ammeter, and a voltmeter. The voltmeter and the ammeter (labeled V and A) can be considered ideal; that is, their resistances are infinity and zero, respectively. The current in the resistor is l, and the voltage across it is V. The internal resistance of the battery r int is not zero.Part AWhat is the ammeter reading l?Express your answer in terms of E, R, and r int.Part BWhat is the voltmeter reading V?Express your answer in terms of E, R, and r int.Part CWhat is the power PR dissipated in the resistor?Express your answer in terms of I and VPart DAgain, what is the power PR dissipated in the resistor. This time, express your answer in terms of one or more of the following variables: l, r int, and R.Part EFor the third time, what is the power PR dissipated in the resistor?Express your answer in terms of one or more of the following variables: E, r int and R.Part FWhat is the total power P total dissipated in the resistive elements of the circuit?Express your answer in terms of one or more of the following variables: E, r int, and R.Part GWhat is the total power P total dissipated in the resistive elements of the circuit, in terms of the EMF E of the battery and the current in the circuit?Express your answer in terms of E and the ammeter current I.

###### FREE Expert Solution

Kirchhoff's voltage law:

$\overline{){\mathbf{\Sigma }}{\mathbf{V}}{\mathbf{=}}{\mathbf{0}}}$

Voltage, V = iR

Part A

Applying Kirchhoff's loop rule in the lower loop in a counterclockwise direction:

$\begin{array}{rcl}\mathbf{IR}\mathbf{+}\mathbit{I}{\mathbf{r}}_{\mathbf{i}\mathbf{n}\mathbf{t}}\mathbf{-}\mathbf{E}& \mathbf{=}& \mathbf{0}\\ \mathbit{I}\mathbf{\left(}\mathbf{R}\mathbf{+}{\mathbf{r}}_{\mathbf{i}\mathbf{n}\mathbf{t}}\mathbf{\right)}& \mathbf{=}& \mathbf{E}\end{array}$

I = E/(R + rint)

The ammeter reading is expressed as E/(R + rint).

Part B

Voltage is given by, V = IR

90% (478 ratings) ###### Problem Details Learning Goal: To understand how to compute power dissipation in a resistive circuit.

(Figure 1) The circuit in the diagram consists of a battery with EMF E, a resistor with resistance R, an ammeter, and a voltmeter. The voltmeter and the ammeter (labeled V and A) can be considered ideal; that is, their resistances are infinity and zero, respectively. The current in the resistor is l, and the voltage across it is V. The internal resistance of the battery r int is not zero.

Part A
What is the ammeter reading l?
Express your answer in terms of E, R, and r int.

Part B
What is the voltmeter reading V?
Express your answer in terms of E, R, and r int.

Part C
What is the power PR dissipated in the resistor?
Express your answer in terms of I and V

Part D
Again, what is the power PR dissipated in the resistor. This time, express your answer in terms of one or more of the following variables: l, r int, and R.

Part E
For the third time, what is the power PR dissipated in the resistor?
Express your answer in terms of one or more of the following variables: E, r int and R.

Part F
What is the total power P total dissipated in the resistive elements of the circuit?
Express your answer in terms of one or more of the following variables: E, r int, and R.

Part G
What is the total power P total dissipated in the resistive elements of the circuit, in terms of the EMF E of the battery and the current in the circuit?
Express your answer in terms of E and the ammeter current I.

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