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: uploaded image ————————————————————— Intro 2 uploaded image

Part A
What is the reading V of the voltmeter?
Express your answer in terms of the EMF 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?
0;; cal{E}
0;; cal{-E}
cal{E};; 0
cal{E};; cal{E}
Part C
What is the reading I of the ammeter?
Express your answer in terms of EMF and R.
Now assume that the battery has a nonzero internal resistance r (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 EMF, r, and R.
Part E
What is the reading of the voltmeter now?
Express your answer in terms of EMF, r, and R.

Answer

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, V = IR 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. Στο = 0 Here,enclosedis 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, V = &-IR 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.

Step 2 of 6

(b) From above provided circuit across points 1 and 2, According to Kirchhoff’s Voltage law, V = &-IR 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.

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, V = &-IR 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.

Step 4 of 6

(c) From Ohm’s law, the expression of the voltage is expresses as follows, V = IR Substitute, for in the above equation V = IR 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.

Step 5 of 6

(d) From Kirchhoff’s Voltage law, the expression of the voltage is expresses as follows, V = &-IR Substitute for in the above expression of the voltage. V = E-Irint From Ohm’s law, the expression of the voltage, V = IR Substitute E-Trink for in the above equation V = IR and solve for. E- Irine = IR
IR + Irink =
1(R+R) = 8
1=
E
(R+R)

Part d

The ammeter reading is equal to(+ X) =1.


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

Step 6 of 6

(e) From Kirchhoff’s Voltage law, the expression of the voltage, V = &-IR Substitute for in the above expression of the voltage, V = E-Irint Substitute (R+rint) for in the equation V = E-Irint and solve for .

V=8- Trening hay
(R+ Pink Pink
(R+R)
ER
R+ Pink

Part e

The voltmeter reading is equal toER
V=
R+ Pink.


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

Answer

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(+ X) =1.

Part e

The voltmeter reading is equal toER
V=
R+ Pink.

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