Advertisements
Advertisements
प्रश्न
Figure below shows two resistors R1 and R2 connected to a battery having an emf of 40V and negligible internal resistance. A voltmeter having a resistance of. 300 Ω is used to measure the potential difference across R1 Find the reading of the voltmeter.
Advertisements
उत्तर
R1 and R3 are in parallel.
∴ Current I in the circuit remains unchanged in R1 and R3 combination and R2
`1/(R') = 1/200 +1/300 =5/600`
`R' = 600/5 = 120 Omega `
∴ REquivalent in circuit = 120 + 880 = 1000 Ω
I, current.in the circuit =`40/1000 A`
∴ V =` 40/1000 xx 120`
⇒ `V = 48/10`
∴ V = 4.8 V
APPEARS IN
संबंधित प्रश्न
Figure shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.
What will be the effect on the position of zero deflection if only the current flowing through the potentiometer wire is increased?
State any one use of a potentiometer.
What are the disadvantages of a potentiometer over a voltmeter?
Which of the following instruments is not a direct reading instrument?
A battery is connected with a potentiometer wire. The internal resistance of the battery is negligible. If the length of the potentiometer wire of the same material and radius is doubled then ______.
The instrument among the following which measures the e.m.f of a cell most accurately is ______
As a cell age, its internal resistance increases. A voltmeter of resistance 270 Ω connected across an old dry cell reads 1.44 V. However, a potentiometer at the balance point gives a voltage measurement of the cell as 1.5 V. Internal resistance of the cell is ______ Ω.
What is the internal resistance of the cell?
