Home
Class 12
PHYSICS
Figure below shows two resistors R1 and ...

Figure below shows two resistors `R_1 and R_2` connected to a battery having an emf of 40 V and negligible internal resistance. A voltmeter having a resistance of 300 `Omega` is used to measure potential difference across `R_1` Find the reading of the voltmeter.

Text Solution

Verified by Experts


As `R_1` and voltmeter are in parallel, so their equivalent resistance is given by
`1/(R.)=1/(200)+1/(300)`
`implies R. =(200xx300)/(200+300)`
`= (200xx300)/500`
`implies R. = 120Omega`
Now, R. and `R_2` are in series, so their equivalent resistance is given by
`R = R. +880`
` = 120 +880 = 1000 Omega`
Since the current in series combination remains the same.
`:.` Current, `I =V/R`
`=(40)/1000=1/25A`
`:.` Potential drop across the voltmeter
`V=IR.`
`=1/25xx120`
`= 4.8 V `
Promotional Banner

Similar Questions

Explore conceptually related problems

Two resistance of 400 Omega and 800 Omega are connected in series with 6 volt battery of negligible internal resistance. A voltmeter of resistance 10,000 Omega is used to measure the potential difference across 400 Omega . The error in measurement of potential difference in volts approximatley is

In the circuit shows in Fig. 6.42 , resistors X and Y , each with resistance R , are connected to a 6 V battery of negligible internal resistance. A voltmeter, also of resistance R , is connected across Y . What is the reading of the voltmeter?

A voltmeter with resistance 500 Omega is used to measure the emf of a cell of internal resistance 4 Omega . The percentage error in the reading of the voltmeter will be

A cell having an emf E and internal resistance r is connected across a variable external resistance R. As the resistance R is increased, the plot of potential difference V across R is given by

2A cell having an emf epsilon and internal resistance r is connected across a variable external resistance R . As the resistance R is increased, the plot of potential difference V across R is given by

A 5 V battery of negligible internal resistance is connected across a 200 V battery and a resistance of 39 Omega as shown in the figure. Find the value of the current.

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?

Consider the diagram shown below. A Voltmeter of resistance 150 Omega is connected across A and B. The potential drop across B and C measured by voltmeter is

The figure shows a source (a battery) with an emf E of 12 V with an internal resistance r of 2Omega and an external resistance of 4Omega is added to complete the circuit. What are the voltmeter and ammeter readings

A cell of emf 3.4 V and internal resistance 3 Omega is connected to an ammeter having resistance 2Omega and to an external resistance of 100Omega . When a voltmeter is connected across the 100 Omega resistance, the ammeter reading is 0.04 A . Find the voltage reading by the voltmeter and its resistance. Had the voltmeter been an ideal one what would have been its reading?