Home
Class 12
PHYSICS
A voltmeter reads the potential differen...

A voltmeter reads the potential difference across the terminals of an old battery as `1.40 V`, while a potentiometer reads its voltage to be `1.55 V`. The voltmeter resistance is `280 Omega`.

A

`60Omega`

B

`45Omega`

C

`35Omega`

D

`30Omega`

Text Solution

Verified by Experts

The correct Answer is:
4
Let `epsilon` be emf and r be internal resistance of the cell when tno current drawn from the cell, the potential difference b etween its terminals is the emf of the cell.

It is this quantity which the potentionmeter measures.
Therefore, emf of the cell `epsilon=1.55V`
When the `280Omega` voltmeter is connected ac ross the cell as shown in adjacent figure, current flows from the cell.
The potential difference between its terminals is just the voltmeter reading. 140V.
`terefore 1xx280=1.40`
`1=(1.4)/(280)=5xx10^(-3)A`
Also `I=(epsilon)/(280+r)=(1.55)/(28+r)`
`5xx10^(-3)=(1.55)/(28+r)`
`1.4+5xx10^(-3)r=1.55`
`5xx10^(-3)r=0.15`
`r=(0.15)/(5xx10^(-3)=30Omega`
According to m aximum power theoram.
Promotional Banner

Similar Questions

Explore conceptually related problems

A battery of emf 1.4 V and internal resistance 2 Omega is connected to a resistor of 100 omega resistance through an ammeter. This resistance of the ammeter is 4//3 Omega . A voltmeter has also been connected to find the potential difference across the resistor. a. Draw the circuit diagram. b. The ammeter reads 0.02 A . What is the resistance of the voltmeter? c. The voltmeter reads 1.1 V . What is the error in the reading?

In the experiment of calibration of voltmeter, a standard cell of emf 1.1 V is balanced against 440 cm of potentiometer wire. The potentail difference across the ends of a resistance is found to balance aginst 220 cm of the wire. The corresponding reading of the voltmeter is 0.5 V . Find theerror in the reading of voltmeter.

A high resistance voltmeter measures the potential difference across a battery to be 9-0 V. On connecting a 24 2 resistor across the terminals of the battery, the voltmeter reads 7.2 V. Calculate the internal resistance of the battery.

A voltmeter having a resistance of 1800 Omega employed to measure the potential difference across a 200 Omega resistor which is connected to the terminals of a dc power supply having an emf of 50 V and an internal resistance of 20 Omega . What is the percentage decrease in the potential difference across the 200 Omega resistor as a result of connecting the voltmeter across it?

A voltmeter and an ammeter are connected in series to an ideal cell of emf E . The voltmeter reading is V , and the ammeter readings is I . Then (i) V lt E (ii) the voltmeter resistance is V// I (iii) the potential difference across the ammeter is E - V (iv) Voltmeter resistance + ammeter resistance = E//I Correct statements are

If the reading of coltmenter V_1 is 40 V, what is the reading of voltmeter V_2 ?

An ammeter and a voltmeter are connected in series to an ideal battery of emf E. If the ammeter reads I and voltmeter reading V, then

In the adjoining circuit diagram, the readings of ammeter and voltmeter are 2 A and 120 V, respectively. If the value of R is 75Omega , then the voltmeter resistance will be

If the voltage between terminals A and B is 20 V and Zener breakdown voltage is 7 V then the potential difference across 6 Omega resistance is

A capacitor of capacitance 1muF is connected in parallel with a resistance 10Omega and the combination is connected across the terminals of a battery of EMF 50 V and internal resistance 1Omega . The potential difference across the capacitor at steady state is (in Volt)