Advertisements
Advertisements
Question
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.
Advertisements
Solution
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
APPEARS IN
RELATED QUESTIONS
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 battery of e.m.f. 15 V and internal resistance 3 ohm is connected to two resistors of resistances 3 ohm and 6 ohm is series Find:
(a) the current through the battery
(b) the p.d. between the terminals of the battery.
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:
- The reading of the ammeter,
- The potential difference across the terminals of the cells, and
- 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.
When a resistance of 3Ω is connected across a cell, the current flowing is 0.5 A. On changing the resistance to 7Ω, the current becomes 0.25A. Calculate the e.m.f. and the internal resistance of the cell.
(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.
Explain the meaning of the term internal resistance of a cell.
