हिंदी

A battery of e.m.f. 6.0 V supplies current through a circuit in which the resistance can be changed. A high resistance voltmeter is connected across the battery. When the current is 3 A, the voltmeter

Advertisements
Advertisements

प्रश्न

A battery of e.m.f. 6.0 V supplies current through a circuit in which the resistance can be changed. A high resistance voltmeter is connected across the battery. When the current is 3 A, the voltmeter reads 5.4 V. Find the internal resistance of the battery.

संख्यात्मक
Advertisements

उत्तर

Since a battery's terminal potential difference is less than its e.m.f., an increase in circuit current causes the voltmeter reading to fall.

Now E = 6.0 V,

V = 5.4 V,

I = 3.0 A

Internal resistance r = `("E" - "V")/"I"`

= `(6 - 5.4)/3.0`

= `0.3/1.5`

r = 0.2 Ω

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Current Electricity - EXERCISE - 8(B) [पृष्ठ २०१]

APPEARS IN

सेलिना Concise Physics [English] Class 10 ICSE
अध्याय 8 Current Electricity
EXERCISE - 8(B) | Q 2. | पृष्ठ २०१

संबंधित प्रश्न

A cell of Emf 2 V and internal resistance 1.2 Ω is connected with an ammeter of resistance 0.8 Ω and two resistors of 4.5 Ω and 9 Ω as shown in the diagram below:

1) What would be the reading on the Ammeter?

2) What is the potential difference across the terminals of the cell?


Explain why the p.d across the terminals of a cell is more in an open circuit and reduced in a closed circuit. 


A cell of e.m.f. 1.8V and internal resistance 2Ω is connected in series with an ammeter of resistance 0.7Ω and a resistor of 4.5Ω as shown in Fig. 

  1. What would be the reading of the ammeter?
  2. What is the potential difference across the terminals of the cell? 

A cell of e.m.f. ε and internal resistance 𝔯 sends current 1.0 A when it is connected to an external resistance 1.9 Ω. But it sends current 0.5 A when it is connected to an external resistance 3.9 Ω. Calculate the values of ε and 𝔯.


A cell of e.m.f. 2 V and internal resistance 1.2 Ω is connected to an ammeter of resistance 0.8 Ω and two resistors of 4.5 Ω and 9 Ω as shown in following figure.

Find:

  1. The reading of the ammeter,
  2. The potential difference across the terminals of the cells, and
  3. The potential difference across the 4.5 Ω resistor.

A cell of emf. 1.5 V and internal resistance 10 ohms is connected to a resistor of 5 ohms, with an ammeter in series see fig.. What is the reading of the ammeter?


Define the e.m.f. (E) of a cell and the potential difference (V) of a resistor R in terms of the work done in moving a unit charge. State the relation between these two works and the work done in moving a unit charge through a cell connected across the resistor. Take the internal resistance of the cell as ‘r’. Hence obtain an expression for the current i in the circuit.


(a) Calculate the total resistance across AB.

(b) If a cell of e.m.f 2.4 V with negligible internal resistance is connected across AB then calculate the current drawn from the cell.


Study the diagram:

  1. Calculate the total resistance of the circuit.
  2. Calculate the current drawn from the cell.
  3. State whether the current through 10 Ω resistor is greater than, less than or equal to the current through the 12 Ω resistor.

The diagram in Figure shows a cell of e.m.f. ε = 4 volt and internal resistance r = 2 ohm connected to an external resistance R = 8 ohm. The ammeter A measures the current in the circuit and the voltmeter V measures the terminal voltage across the cell. What will be the readings of the ammeter and voltmeter when

  1. the key K is open, and
  2. the key K is closed


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×