# What is the reading v of the voltmeter? express your answer in terms of the emf e.

Consider the circuit shown. (Intro 1 figure) All wires are
considered ideal; that is, they have zero resistance. We will
assume for now that all other elements of the circuit are ideal,
too: The value of resistance is a constant, the internal
resistances of the battery and the ammeter are zero, and the
internal resistance of the voltmeter is infinitely large. Intro 1: ————————————————————— Intro 2 Part A What is the reading of the voltmeter? Express your answer in terms of the
EMF . Part B
The voltmeter, as can be seen in the figure,
is connected to points 1 and 3. What are the respective voltage
differences between points 1 and 2 and between points 2 and
3?

Part C
What is the reading of the ammeter?

and .

Now assume that the battery has a
nonzero internal resistance (but the voltmeter and the ammeter remain ideal).
(Intro 2 figure)
Part D
What is the reading of the ammeter now?
, , and .

Part E
What is the reading of the voltmeter
now?
, , and .        Part B The voltmeter, as can be seen in the figure,
is connected to points 1 and 3. What are the respective voltage
differences between points 1 and 2 and between points 2 and
3?

Part C
What is the reading of the ammeter?

and .

Now assume that the battery has a
nonzero internal resistance (but the voltmeter and the ammeter remain ideal).
(Intro 2 figure)
Part D
What is the reading of the ammeter now?
, , and .

Part E
What is the reading of the voltmeter
now?
, , and .

Part C
What is the reading of the ammeter?

and .

Now assume that the battery has a
nonzero internal resistance (but the voltmeter and the ammeter remain ideal).
(Intro 2 figure)
Part D
What is the reading of the ammeter now?
, , and .

Part E
What is the reading of the voltmeter
now?
, , and .

Part C
What is the reading of the ammeter?

and .

Now assume that the battery has a
nonzero internal resistance (but the voltmeter and the ammeter remain ideal).
(Intro 2 figure)
Part D
What is the reading of the ammeter now?
, , and .

Part E
What is the reading of the voltmeter
now?
, , and .

Part C
What is the reading of the ammeter?

and .

Now assume that the battery has a
nonzero internal resistance (but the voltmeter and the ammeter remain ideal).
(Intro 2 figure)
Part D
What is the reading of the ammeter now?
, , and .

Part E
What is the reading of the voltmeter
now?
, , and .

Part C
What is the reading of the ammeter?

and .

Now assume that the battery has a
nonzero internal resistance (but the voltmeter and the ammeter remain ideal).
(Intro 2 figure)
Part D
What is the reading of the ammeter now?
, , and .

Part E
What is the reading of the voltmeter
now?
, , and .

Part C
What is the reading of the ammeter?

and .

Now assume that the battery has a
nonzero internal resistance (but the voltmeter and the ammeter remain ideal).
(Intro 2 figure)
Part D
What is the reading of the ammeter now?
, , and .

Part E
What is the reading of the voltmeter
now?
, , and .

Part C
What is the reading of the ammeter?

and .

Now assume that the battery has a
nonzero internal resistance (but the voltmeter and the ammeter remain ideal).
(Intro 2 figure)
Part D
What is the reading of the ammeter now?
, , and .

Part E
What is the reading of the voltmeter
now?
, , and .

Part C
What is the reading of the ammeter?

and .

Now assume that the battery has a
nonzero internal resistance (but the voltmeter and the ammeter remain ideal).
(Intro 2 figure)
Part D
What is the reading of the ammeter now?
, , and .

Part E
What is the reading of the voltmeter
now?
, , and .

Part C
What is the reading of the ammeter?

and .

Now assume that the battery has a
nonzero internal resistance (but the voltmeter and the ammeter remain ideal).
(Intro 2 figure)
Part D
What is the reading of the ammeter now?
, , and .

Part E
What is the reading of the voltmeter
now?
, , and .

Part C
What is the reading of the ammeter?

and .

Now assume that the battery has a
nonzero internal resistance (but the voltmeter and the ammeter remain ideal).
(Intro 2 figure)
Part D
What is the reading of the ammeter now?
, , and .

Part E
What is the reading of the voltmeter
now?
, , and .

Part C
What is the reading of the ammeter?

and .

Now assume that the battery has a
nonzero internal resistance (but the voltmeter and the ammeter remain ideal).
(Intro 2 figure)
Part D
What is the reading of the ammeter now?
, , and .

Part E
What is the reading of the voltmeter
now?
, , and .

Part C What is the reading of the ammeter?
and .

Now assume that the battery has a
nonzero internal resistance (but the voltmeter and the ammeter remain ideal).
(Intro 2 figure)
Part D
What is the reading of the ammeter now?
, , and .

and .
Now assume that the battery has a
nonzero internal resistance (but the voltmeter and the ammeter remain ideal).
(Intro 2 figure)
Part D
What is the reading of the ammeter now?
, , and .
Now assume that the battery has a
nonzero internal resistance (but the voltmeter and the ammeter remain ideal).
(Intro 2 figure) Part D What is the reading of the ammeter now? Express your answer in terms of
, , and .
Part E
What is the reading of the voltmeter
now?
, , and .
Part E What is the reading of the voltmeter
now? Express your answer in terms of
, , and .

General guidance

Concepts and reason
The question can be solved by using the Ohm’s law and Kirchhoff’s law, which relates voltage, current and resistance.
A battery or energy source generates an in a closed circuit. The potential difference or causes current to flow in the circuit. The emf is dropped across the resistor. These parameters are related in a single equation given by Ohm’s law.

Fundamentals

From Ohm’s law, the expression of the voltage is expresses as follows,

Here, is the voltage or potential difference, is the current and is the resistance.
From the definition of the Kirchhoff’s current law: The algebraic Sum of the currents in a network of conductor meeting at a point is zero.

Here,is the enclosed current.
From the definition of Kirchhoff’s Voltage law, the algebraic sum of the products of the resistance of the conductors and the current in them in a closed loop is equal to the total available in that loop.

Step-by-step

Step 1 of 6

(a)
From Kirchhoff’s Voltage law, the total voltage across the circuit is expresses as follows,

Here, is the induced.
Since, across 1 and 3 and as no current flows through voltmeter because of high resistance.

Since, voltmeter is connected across 1 and 3 it must read

Part a
The voltmeter reading is equal to.

No current flows through voltmeter because of high resistance. Therefore, voltmeter does not alter the. As the potential resistance that can alter is absent, voltmeter reads a value equal toof the battery.

Kirchhoff’s law must be adopted carefully considering the signs of the terms involved. Therefore, use the correct expression of the Kirchhoff’s voltage in the closed circuit.

Use Kirchhoff’s Voltage law to find the voltage across the given points.

Step 2 of 6

(b)
From above provided circuit across points 1 and 2, According to Kirchhoff’s Voltage law,

Across 1 and 2, and .
Therefore, the voltage is equal to zero.

Part b
Voltage difference between point 1 and 2: .

Since there is no resistance from point 1 to 2, both these points are at the same potential and the voltage difference between them is zero.

Use Kirchhoff’s Voltage law to find the voltage across the given points.

Step 3 of 6

From the above provided diagram, the voltage across points 2 and 3 is expresses as follows:
From Kirchhoff’s Voltage law, the voltage across the circuit is expresses as follows,

Since, across points 2 and 3 is expresses as follows:

Part b
Voltage difference between point 2 and 3: .

Since there is a battery from point 2 to 3, thus the potential between these points is emf of the battery.

The question asks voltage difference between points and not to be confused with voltage at the point.

Use Ohm’s law to find the current.

Step 4 of 6

(c)
From Ohm’s law, the expression of the voltage is expresses as follows,

Substitute, for in the above equation and solve for .

Part c
The current is equal to, .

As resistance is connected between point 1 and 3, the voltage difference between these points is.

Do not use the expression to the calculation of the current. Therefore, use the correct expression for the current.

Use Ohm’s law to find the current.

Step 5 of 6

(d)
From Kirchhoff’s Voltage law, the expression of the voltage is expresses as follows,

Substitute for in the above expression of the voltage.

From Ohm’s law, the expression of the voltage,

Substitute for in the above equation and solve for.

Part d
The ammeter reading is equal to.

The voltage of the battery drops because of its internal resistance. The voltage drop is given by Ohm’s law.

The voltage difference between is not anymore, it has changed due to internal resistance of the battery. Therefore, use the correct expression of the voltage for the calculation of the current.

Use Kirchhoff’s Voltage law to find the new voltage.

Step 6 of 6

(e)
From Kirchhoff’s Voltage law, the expression of the voltage,

Substitute for in the above expression of the voltage,

Substitute for in the equation and solve for .

Part e
The voltmeter reading is equal to.

The voltage of the battery drops because of its internal resistance. The voltage drop is given by Ohm’s law.

Do not use the expression of the voltage to the calculation of the voltage. Therefore use the correct expression of the voltage to the calculation of the voltage.

Part a
The voltmeter reading is equal to.

Part b
Voltage difference between point 1 and 2: .

Part b
Voltage difference between point 2 and 3: .

Part c
The current is equal to, .

Part d
The ammeter reading is equal to.

Part e
The voltmeter reading is equal to.

Part a
The voltmeter reading is equal to.

Part b
Voltage difference between point 1 and 2: .

Part b
Voltage difference between point 2 and 3: .

Part c
The current is equal to, .

Part d
The ammeter reading is equal to.

Part e
The voltmeter reading is equal to.

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Στο = 0
enclosed
V = &-IR
V = &-IR
V = &-IR
V = IR
V = IR
V = &-IR
V = E-Irint
V = IR
E-Trink
V = IR
E- Irine = IR IR + Irink = 1(R+R) = 8 1= E (R+R)
V = &-IR
V = E-Irint
(R+rint)
V = E-Irint
V=8- Trening hay (R+ Pink Pink (R+R) ER R+ Pink
V = 2€
V=-&
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